How To solve This Perfect Square Word Problem












3












$begingroup$


Here's a problem about perfect squares and it's very hard for me. I tried to solve but I got stuck.



Last year, the town of Whipple had a population that was a perfect square. Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square. Next month, 100 more people will move to Whipple, making the population a perfect square again. What was the original population of Whipple?



Here's what I did:



Let the population last year be n, so n = x^2 and x = √n
Last month: n + 100 = x^2 + 1
Next Month: n + 200 = x^2 ...



and i Got stuck there. I don't know where I am going ... Your help is appreciated






algebra-precalculus square-numbers







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    Here's a problem about perfect squares and it's very hard for me. I tried to solve but I got stuck.



    Last year, the town of Whipple had a population that was a perfect square. Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square. Next month, 100 more people will move to Whipple, making the population a perfect square again. What was the original population of Whipple?



    Here's what I did:



    Let the population last year be n, so n = x^2 and x = √n
    Last month: n + 100 = x^2 + 1
    Next Month: n + 200 = x^2 ...



    and i Got stuck there. I don't know where I am going ... Your help is appreciated






    algebra-precalculus square-numbers







    share|cite|improve this question









    $endgroup$















      3












      3








      3


      1



      $begingroup$


      Here's a problem about perfect squares and it's very hard for me. I tried to solve but I got stuck.



      Last year, the town of Whipple had a population that was a perfect square. Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square. Next month, 100 more people will move to Whipple, making the population a perfect square again. What was the original population of Whipple?



      Here's what I did:



      Let the population last year be n, so n = x^2 and x = √n
      Last month: n + 100 = x^2 + 1
      Next Month: n + 200 = x^2 ...



      and i Got stuck there. I don't know where I am going ... Your help is appreciated






      algebra-precalculus square-numbers







      share|cite|improve this question









      $endgroup$




      Here's a problem about perfect squares and it's very hard for me. I tried to solve but I got stuck.



      Last year, the town of Whipple had a population that was a perfect square. Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square. Next month, 100 more people will move to Whipple, making the population a perfect square again. What was the original population of Whipple?



      Here's what I did:



      Let the population last year be n, so n = x^2 and x = √n
      Last month: n + 100 = x^2 + 1
      Next Month: n + 200 = x^2 ...



      and i Got stuck there. I don't know where I am going ... Your help is appreciated






      algebra-precalculus square-numbers




      algebra-precalculus square-numbers






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 29 at 23:38









      harpey1111harpey1111

      656




      656






















          4 Answers
          4






          active

          oldest

          votes


















          7












          $begingroup$

          I like where Will Jagy starts.



          $$x^2+99 = y^2\
          x^2 + 200 = z^2$$



          to continue I would subtract one from the other



          $$z^2 - y^2 = 101\(z+y)(z-y) = 101$$



          $101$ is prime



          $$z+y = 101\z-y = 1\z = 51\y = 50\x = sqrt{51^2 - 200} = 49$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            good...........
            $endgroup$
            – Will Jagy
            Jan 30 at 0:43










          • $begingroup$
            on this part, I just don't understand how you got the z - y = 100 and why you need to subtract the two equations. :(
            $endgroup$
            – harpey1111
            Jan 30 at 22:34










          • $begingroup$
            and maybe that 50^2 is supposed tobe 51^2 right?
            $endgroup$
            – harpey1111
            Jan 30 at 22:45










          • $begingroup$
            @harpey1111 I subtracted one from the other, for a couple of reasons. One, it eliminates the x^2 term. Two, it creates a prime number to work with. And yes, $51^2 - 200 = 49^2$ thanks for pointing that out.
            $endgroup$
            – Doug M
            Jan 30 at 23:08










          • $begingroup$
            thank you so much for your help. I got the problem right. have a wonderful night :)
            $endgroup$
            – harpey1111
            Jan 30 at 23:34



















          10












          $begingroup$

          Be careful with your variable names; it will clarify things for you. You've used $x$ to mean three different things!



          Let $n=x^2$; then $n+200=y^2$, so that $x^2+200 = y^2$. Rewriting gives $y^2-x^2=(y-x)(y+x)=200$. Can you make progress from there?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I kind of get it now but I still don't know how to use it to solve the problem because the "Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square." makes me really confused. I'm so sorry.
            $endgroup$
            – harpey1111
            Jan 29 at 23:57










          • $begingroup$
            thank you. I got the problem right . I really appreciate your help. and I'll take note of your advice. it's really helpful.
            $endgroup$
            – harpey1111
            Jan 30 at 23:38










          • $begingroup$
            You're welcome. You should accept, by clicking the green checkmark, whichever of the several great answers you like best.
            $endgroup$
            – rogerl
            Jan 31 at 14:06



















          7












          $begingroup$

          You have a square $x^2$ and you know that $x^2+99=y^2$ and $y^2+101=z^2$.



          It is well known that the sum of the first $n$ odd numbers equals $n^2$, meaning that adding sequential odd numbers to a smaller square yields the squares of the next larger numbers, i.e. if the first $n$ odd numbers sum to $n^2$ then the first $n+1$ odd numbers sum to $(n+1)^2$, etc.



          Now notice that $99$ and $101$ are sequential odd numbers. A little manipulation reveals that $99$ is the $50$th odd number ($99=(2cdot 50)-1)$) and $101$ is the $51$st odd number ($101=(2cdot 51)-1)$). The starting square in this scenario would be the square which is the sum of the first $49$ odd numbers. So the squares that solve the problem are $49^2$, $50^2$, and $51^2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you for spending some of your time explaining this. It helped me. I really appreciate it.
            $endgroup$
            – harpey1111
            Jan 30 at 23:39



















          4












          $begingroup$

          $$ x^2 + 99 = y^2 $$
          $$ x^2 + 200 = z^2 $$
          $$ (y+x)(y-x) = 99 = 99 cdot 1 = 33 cdot 3 = 11 cdot 9$$
          The choices for $x$ are
          $$ frac{99-1}{2} , ; ; ; frac{33-3}{2} , ; ; ; frac{11-9}{2} , ; ; ; $$
          Then confirm that $x^2 + 99$ is really a square, and check whether $x^2 + 200$ is a square, for the three possible $x$ values.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            how did you get "divided by 2"?
            $endgroup$
            – harpey1111
            Jan 30 at 0:32










          • $begingroup$
            @harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$
            $endgroup$
            – Will Jagy
            Jan 30 at 0:37










          • $begingroup$
            Thank you so much. I got the problem right. thanks a lot. Have a wonderful night :)
            $endgroup$
            – harpey1111
            Jan 30 at 23:40











          Your Answer





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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          7












          $begingroup$

          I like where Will Jagy starts.



          $$x^2+99 = y^2\
          x^2 + 200 = z^2$$



          to continue I would subtract one from the other



          $$z^2 - y^2 = 101\(z+y)(z-y) = 101$$



          $101$ is prime



          $$z+y = 101\z-y = 1\z = 51\y = 50\x = sqrt{51^2 - 200} = 49$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            good...........
            $endgroup$
            – Will Jagy
            Jan 30 at 0:43










          • $begingroup$
            on this part, I just don't understand how you got the z - y = 100 and why you need to subtract the two equations. :(
            $endgroup$
            – harpey1111
            Jan 30 at 22:34










          • $begingroup$
            and maybe that 50^2 is supposed tobe 51^2 right?
            $endgroup$
            – harpey1111
            Jan 30 at 22:45










          • $begingroup$
            @harpey1111 I subtracted one from the other, for a couple of reasons. One, it eliminates the x^2 term. Two, it creates a prime number to work with. And yes, $51^2 - 200 = 49^2$ thanks for pointing that out.
            $endgroup$
            – Doug M
            Jan 30 at 23:08










          • $begingroup$
            thank you so much for your help. I got the problem right. have a wonderful night :)
            $endgroup$
            – harpey1111
            Jan 30 at 23:34
















          7












          $begingroup$

          I like where Will Jagy starts.



          $$x^2+99 = y^2\
          x^2 + 200 = z^2$$



          to continue I would subtract one from the other



          $$z^2 - y^2 = 101\(z+y)(z-y) = 101$$



          $101$ is prime



          $$z+y = 101\z-y = 1\z = 51\y = 50\x = sqrt{51^2 - 200} = 49$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            good...........
            $endgroup$
            – Will Jagy
            Jan 30 at 0:43










          • $begingroup$
            on this part, I just don't understand how you got the z - y = 100 and why you need to subtract the two equations. :(
            $endgroup$
            – harpey1111
            Jan 30 at 22:34










          • $begingroup$
            and maybe that 50^2 is supposed tobe 51^2 right?
            $endgroup$
            – harpey1111
            Jan 30 at 22:45










          • $begingroup$
            @harpey1111 I subtracted one from the other, for a couple of reasons. One, it eliminates the x^2 term. Two, it creates a prime number to work with. And yes, $51^2 - 200 = 49^2$ thanks for pointing that out.
            $endgroup$
            – Doug M
            Jan 30 at 23:08










          • $begingroup$
            thank you so much for your help. I got the problem right. have a wonderful night :)
            $endgroup$
            – harpey1111
            Jan 30 at 23:34














          7












          7








          7





          $begingroup$

          I like where Will Jagy starts.



          $$x^2+99 = y^2\
          x^2 + 200 = z^2$$



          to continue I would subtract one from the other



          $$z^2 - y^2 = 101\(z+y)(z-y) = 101$$



          $101$ is prime



          $$z+y = 101\z-y = 1\z = 51\y = 50\x = sqrt{51^2 - 200} = 49$$






          share|cite|improve this answer











          $endgroup$



          I like where Will Jagy starts.



          $$x^2+99 = y^2\
          x^2 + 200 = z^2$$



          to continue I would subtract one from the other



          $$z^2 - y^2 = 101\(z+y)(z-y) = 101$$



          $101$ is prime



          $$z+y = 101\z-y = 1\z = 51\y = 50\x = sqrt{51^2 - 200} = 49$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 30 at 23:05

























          answered Jan 30 at 0:37









          Doug MDoug M

          45.3k31954




          45.3k31954












          • $begingroup$
            good...........
            $endgroup$
            – Will Jagy
            Jan 30 at 0:43










          • $begingroup$
            on this part, I just don't understand how you got the z - y = 100 and why you need to subtract the two equations. :(
            $endgroup$
            – harpey1111
            Jan 30 at 22:34










          • $begingroup$
            and maybe that 50^2 is supposed tobe 51^2 right?
            $endgroup$
            – harpey1111
            Jan 30 at 22:45










          • $begingroup$
            @harpey1111 I subtracted one from the other, for a couple of reasons. One, it eliminates the x^2 term. Two, it creates a prime number to work with. And yes, $51^2 - 200 = 49^2$ thanks for pointing that out.
            $endgroup$
            – Doug M
            Jan 30 at 23:08










          • $begingroup$
            thank you so much for your help. I got the problem right. have a wonderful night :)
            $endgroup$
            – harpey1111
            Jan 30 at 23:34


















          • $begingroup$
            good...........
            $endgroup$
            – Will Jagy
            Jan 30 at 0:43










          • $begingroup$
            on this part, I just don't understand how you got the z - y = 100 and why you need to subtract the two equations. :(
            $endgroup$
            – harpey1111
            Jan 30 at 22:34










          • $begingroup$
            and maybe that 50^2 is supposed tobe 51^2 right?
            $endgroup$
            – harpey1111
            Jan 30 at 22:45










          • $begingroup$
            @harpey1111 I subtracted one from the other, for a couple of reasons. One, it eliminates the x^2 term. Two, it creates a prime number to work with. And yes, $51^2 - 200 = 49^2$ thanks for pointing that out.
            $endgroup$
            – Doug M
            Jan 30 at 23:08










          • $begingroup$
            thank you so much for your help. I got the problem right. have a wonderful night :)
            $endgroup$
            – harpey1111
            Jan 30 at 23:34
















          $begingroup$
          good...........
          $endgroup$
          – Will Jagy
          Jan 30 at 0:43




          $begingroup$
          good...........
          $endgroup$
          – Will Jagy
          Jan 30 at 0:43












          $begingroup$
          on this part, I just don't understand how you got the z - y = 100 and why you need to subtract the two equations. :(
          $endgroup$
          – harpey1111
          Jan 30 at 22:34




          $begingroup$
          on this part, I just don't understand how you got the z - y = 100 and why you need to subtract the two equations. :(
          $endgroup$
          – harpey1111
          Jan 30 at 22:34












          $begingroup$
          and maybe that 50^2 is supposed tobe 51^2 right?
          $endgroup$
          – harpey1111
          Jan 30 at 22:45




          $begingroup$
          and maybe that 50^2 is supposed tobe 51^2 right?
          $endgroup$
          – harpey1111
          Jan 30 at 22:45












          $begingroup$
          @harpey1111 I subtracted one from the other, for a couple of reasons. One, it eliminates the x^2 term. Two, it creates a prime number to work with. And yes, $51^2 - 200 = 49^2$ thanks for pointing that out.
          $endgroup$
          – Doug M
          Jan 30 at 23:08




          $begingroup$
          @harpey1111 I subtracted one from the other, for a couple of reasons. One, it eliminates the x^2 term. Two, it creates a prime number to work with. And yes, $51^2 - 200 = 49^2$ thanks for pointing that out.
          $endgroup$
          – Doug M
          Jan 30 at 23:08












          $begingroup$
          thank you so much for your help. I got the problem right. have a wonderful night :)
          $endgroup$
          – harpey1111
          Jan 30 at 23:34




          $begingroup$
          thank you so much for your help. I got the problem right. have a wonderful night :)
          $endgroup$
          – harpey1111
          Jan 30 at 23:34











          10












          $begingroup$

          Be careful with your variable names; it will clarify things for you. You've used $x$ to mean three different things!



          Let $n=x^2$; then $n+200=y^2$, so that $x^2+200 = y^2$. Rewriting gives $y^2-x^2=(y-x)(y+x)=200$. Can you make progress from there?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I kind of get it now but I still don't know how to use it to solve the problem because the "Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square." makes me really confused. I'm so sorry.
            $endgroup$
            – harpey1111
            Jan 29 at 23:57










          • $begingroup$
            thank you. I got the problem right . I really appreciate your help. and I'll take note of your advice. it's really helpful.
            $endgroup$
            – harpey1111
            Jan 30 at 23:38










          • $begingroup$
            You're welcome. You should accept, by clicking the green checkmark, whichever of the several great answers you like best.
            $endgroup$
            – rogerl
            Jan 31 at 14:06
















          10












          $begingroup$

          Be careful with your variable names; it will clarify things for you. You've used $x$ to mean three different things!



          Let $n=x^2$; then $n+200=y^2$, so that $x^2+200 = y^2$. Rewriting gives $y^2-x^2=(y-x)(y+x)=200$. Can you make progress from there?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I kind of get it now but I still don't know how to use it to solve the problem because the "Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square." makes me really confused. I'm so sorry.
            $endgroup$
            – harpey1111
            Jan 29 at 23:57










          • $begingroup$
            thank you. I got the problem right . I really appreciate your help. and I'll take note of your advice. it's really helpful.
            $endgroup$
            – harpey1111
            Jan 30 at 23:38










          • $begingroup$
            You're welcome. You should accept, by clicking the green checkmark, whichever of the several great answers you like best.
            $endgroup$
            – rogerl
            Jan 31 at 14:06














          10












          10








          10





          $begingroup$

          Be careful with your variable names; it will clarify things for you. You've used $x$ to mean three different things!



          Let $n=x^2$; then $n+200=y^2$, so that $x^2+200 = y^2$. Rewriting gives $y^2-x^2=(y-x)(y+x)=200$. Can you make progress from there?






          share|cite|improve this answer









          $endgroup$



          Be careful with your variable names; it will clarify things for you. You've used $x$ to mean three different things!



          Let $n=x^2$; then $n+200=y^2$, so that $x^2+200 = y^2$. Rewriting gives $y^2-x^2=(y-x)(y+x)=200$. Can you make progress from there?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 29 at 23:44









          rogerlrogerl

          18k22748




          18k22748












          • $begingroup$
            I kind of get it now but I still don't know how to use it to solve the problem because the "Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square." makes me really confused. I'm so sorry.
            $endgroup$
            – harpey1111
            Jan 29 at 23:57










          • $begingroup$
            thank you. I got the problem right . I really appreciate your help. and I'll take note of your advice. it's really helpful.
            $endgroup$
            – harpey1111
            Jan 30 at 23:38










          • $begingroup$
            You're welcome. You should accept, by clicking the green checkmark, whichever of the several great answers you like best.
            $endgroup$
            – rogerl
            Jan 31 at 14:06


















          • $begingroup$
            I kind of get it now but I still don't know how to use it to solve the problem because the "Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square." makes me really confused. I'm so sorry.
            $endgroup$
            – harpey1111
            Jan 29 at 23:57










          • $begingroup$
            thank you. I got the problem right . I really appreciate your help. and I'll take note of your advice. it's really helpful.
            $endgroup$
            – harpey1111
            Jan 30 at 23:38










          • $begingroup$
            You're welcome. You should accept, by clicking the green checkmark, whichever of the several great answers you like best.
            $endgroup$
            – rogerl
            Jan 31 at 14:06
















          $begingroup$
          I kind of get it now but I still don't know how to use it to solve the problem because the "Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square." makes me really confused. I'm so sorry.
          $endgroup$
          – harpey1111
          Jan 29 at 23:57




          $begingroup$
          I kind of get it now but I still don't know how to use it to solve the problem because the "Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square." makes me really confused. I'm so sorry.
          $endgroup$
          – harpey1111
          Jan 29 at 23:57












          $begingroup$
          thank you. I got the problem right . I really appreciate your help. and I'll take note of your advice. it's really helpful.
          $endgroup$
          – harpey1111
          Jan 30 at 23:38




          $begingroup$
          thank you. I got the problem right . I really appreciate your help. and I'll take note of your advice. it's really helpful.
          $endgroup$
          – harpey1111
          Jan 30 at 23:38












          $begingroup$
          You're welcome. You should accept, by clicking the green checkmark, whichever of the several great answers you like best.
          $endgroup$
          – rogerl
          Jan 31 at 14:06




          $begingroup$
          You're welcome. You should accept, by clicking the green checkmark, whichever of the several great answers you like best.
          $endgroup$
          – rogerl
          Jan 31 at 14:06











          7












          $begingroup$

          You have a square $x^2$ and you know that $x^2+99=y^2$ and $y^2+101=z^2$.



          It is well known that the sum of the first $n$ odd numbers equals $n^2$, meaning that adding sequential odd numbers to a smaller square yields the squares of the next larger numbers, i.e. if the first $n$ odd numbers sum to $n^2$ then the first $n+1$ odd numbers sum to $(n+1)^2$, etc.



          Now notice that $99$ and $101$ are sequential odd numbers. A little manipulation reveals that $99$ is the $50$th odd number ($99=(2cdot 50)-1)$) and $101$ is the $51$st odd number ($101=(2cdot 51)-1)$). The starting square in this scenario would be the square which is the sum of the first $49$ odd numbers. So the squares that solve the problem are $49^2$, $50^2$, and $51^2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you for spending some of your time explaining this. It helped me. I really appreciate it.
            $endgroup$
            – harpey1111
            Jan 30 at 23:39
















          7












          $begingroup$

          You have a square $x^2$ and you know that $x^2+99=y^2$ and $y^2+101=z^2$.



          It is well known that the sum of the first $n$ odd numbers equals $n^2$, meaning that adding sequential odd numbers to a smaller square yields the squares of the next larger numbers, i.e. if the first $n$ odd numbers sum to $n^2$ then the first $n+1$ odd numbers sum to $(n+1)^2$, etc.



          Now notice that $99$ and $101$ are sequential odd numbers. A little manipulation reveals that $99$ is the $50$th odd number ($99=(2cdot 50)-1)$) and $101$ is the $51$st odd number ($101=(2cdot 51)-1)$). The starting square in this scenario would be the square which is the sum of the first $49$ odd numbers. So the squares that solve the problem are $49^2$, $50^2$, and $51^2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you for spending some of your time explaining this. It helped me. I really appreciate it.
            $endgroup$
            – harpey1111
            Jan 30 at 23:39














          7












          7








          7





          $begingroup$

          You have a square $x^2$ and you know that $x^2+99=y^2$ and $y^2+101=z^2$.



          It is well known that the sum of the first $n$ odd numbers equals $n^2$, meaning that adding sequential odd numbers to a smaller square yields the squares of the next larger numbers, i.e. if the first $n$ odd numbers sum to $n^2$ then the first $n+1$ odd numbers sum to $(n+1)^2$, etc.



          Now notice that $99$ and $101$ are sequential odd numbers. A little manipulation reveals that $99$ is the $50$th odd number ($99=(2cdot 50)-1)$) and $101$ is the $51$st odd number ($101=(2cdot 51)-1)$). The starting square in this scenario would be the square which is the sum of the first $49$ odd numbers. So the squares that solve the problem are $49^2$, $50^2$, and $51^2$.






          share|cite|improve this answer









          $endgroup$



          You have a square $x^2$ and you know that $x^2+99=y^2$ and $y^2+101=z^2$.



          It is well known that the sum of the first $n$ odd numbers equals $n^2$, meaning that adding sequential odd numbers to a smaller square yields the squares of the next larger numbers, i.e. if the first $n$ odd numbers sum to $n^2$ then the first $n+1$ odd numbers sum to $(n+1)^2$, etc.



          Now notice that $99$ and $101$ are sequential odd numbers. A little manipulation reveals that $99$ is the $50$th odd number ($99=(2cdot 50)-1)$) and $101$ is the $51$st odd number ($101=(2cdot 51)-1)$). The starting square in this scenario would be the square which is the sum of the first $49$ odd numbers. So the squares that solve the problem are $49^2$, $50^2$, and $51^2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 30 at 2:48









          Keith BackmanKeith Backman

          1,4281812




          1,4281812












          • $begingroup$
            thank you for spending some of your time explaining this. It helped me. I really appreciate it.
            $endgroup$
            – harpey1111
            Jan 30 at 23:39


















          • $begingroup$
            thank you for spending some of your time explaining this. It helped me. I really appreciate it.
            $endgroup$
            – harpey1111
            Jan 30 at 23:39
















          $begingroup$
          thank you for spending some of your time explaining this. It helped me. I really appreciate it.
          $endgroup$
          – harpey1111
          Jan 30 at 23:39




          $begingroup$
          thank you for spending some of your time explaining this. It helped me. I really appreciate it.
          $endgroup$
          – harpey1111
          Jan 30 at 23:39











          4












          $begingroup$

          $$ x^2 + 99 = y^2 $$
          $$ x^2 + 200 = z^2 $$
          $$ (y+x)(y-x) = 99 = 99 cdot 1 = 33 cdot 3 = 11 cdot 9$$
          The choices for $x$ are
          $$ frac{99-1}{2} , ; ; ; frac{33-3}{2} , ; ; ; frac{11-9}{2} , ; ; ; $$
          Then confirm that $x^2 + 99$ is really a square, and check whether $x^2 + 200$ is a square, for the three possible $x$ values.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            how did you get "divided by 2"?
            $endgroup$
            – harpey1111
            Jan 30 at 0:32










          • $begingroup$
            @harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$
            $endgroup$
            – Will Jagy
            Jan 30 at 0:37










          • $begingroup$
            Thank you so much. I got the problem right. thanks a lot. Have a wonderful night :)
            $endgroup$
            – harpey1111
            Jan 30 at 23:40
















          4












          $begingroup$

          $$ x^2 + 99 = y^2 $$
          $$ x^2 + 200 = z^2 $$
          $$ (y+x)(y-x) = 99 = 99 cdot 1 = 33 cdot 3 = 11 cdot 9$$
          The choices for $x$ are
          $$ frac{99-1}{2} , ; ; ; frac{33-3}{2} , ; ; ; frac{11-9}{2} , ; ; ; $$
          Then confirm that $x^2 + 99$ is really a square, and check whether $x^2 + 200$ is a square, for the three possible $x$ values.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            how did you get "divided by 2"?
            $endgroup$
            – harpey1111
            Jan 30 at 0:32










          • $begingroup$
            @harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$
            $endgroup$
            – Will Jagy
            Jan 30 at 0:37










          • $begingroup$
            Thank you so much. I got the problem right. thanks a lot. Have a wonderful night :)
            $endgroup$
            – harpey1111
            Jan 30 at 23:40














          4












          4








          4





          $begingroup$

          $$ x^2 + 99 = y^2 $$
          $$ x^2 + 200 = z^2 $$
          $$ (y+x)(y-x) = 99 = 99 cdot 1 = 33 cdot 3 = 11 cdot 9$$
          The choices for $x$ are
          $$ frac{99-1}{2} , ; ; ; frac{33-3}{2} , ; ; ; frac{11-9}{2} , ; ; ; $$
          Then confirm that $x^2 + 99$ is really a square, and check whether $x^2 + 200$ is a square, for the three possible $x$ values.






          share|cite|improve this answer











          $endgroup$



          $$ x^2 + 99 = y^2 $$
          $$ x^2 + 200 = z^2 $$
          $$ (y+x)(y-x) = 99 = 99 cdot 1 = 33 cdot 3 = 11 cdot 9$$
          The choices for $x$ are
          $$ frac{99-1}{2} , ; ; ; frac{33-3}{2} , ; ; ; frac{11-9}{2} , ; ; ; $$
          Then confirm that $x^2 + 99$ is really a square, and check whether $x^2 + 200$ is a square, for the three possible $x$ values.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 30 at 0:26

























          answered Jan 30 at 0:20









          Will JagyWill Jagy

          104k5102201




          104k5102201












          • $begingroup$
            how did you get "divided by 2"?
            $endgroup$
            – harpey1111
            Jan 30 at 0:32










          • $begingroup$
            @harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$
            $endgroup$
            – Will Jagy
            Jan 30 at 0:37










          • $begingroup$
            Thank you so much. I got the problem right. thanks a lot. Have a wonderful night :)
            $endgroup$
            – harpey1111
            Jan 30 at 23:40


















          • $begingroup$
            how did you get "divided by 2"?
            $endgroup$
            – harpey1111
            Jan 30 at 0:32










          • $begingroup$
            @harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$
            $endgroup$
            – Will Jagy
            Jan 30 at 0:37










          • $begingroup$
            Thank you so much. I got the problem right. thanks a lot. Have a wonderful night :)
            $endgroup$
            – harpey1111
            Jan 30 at 23:40
















          $begingroup$
          how did you get "divided by 2"?
          $endgroup$
          – harpey1111
          Jan 30 at 0:32




          $begingroup$
          how did you get "divided by 2"?
          $endgroup$
          – harpey1111
          Jan 30 at 0:32












          $begingroup$
          @harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$
          $endgroup$
          – Will Jagy
          Jan 30 at 0:37




          $begingroup$
          @harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$
          $endgroup$
          – Will Jagy
          Jan 30 at 0:37












          $begingroup$
          Thank you so much. I got the problem right. thanks a lot. Have a wonderful night :)
          $endgroup$
          – harpey1111
          Jan 30 at 23:40




          $begingroup$
          Thank you so much. I got the problem right. thanks a lot. Have a wonderful night :)
          $endgroup$
          – harpey1111
          Jan 30 at 23:40


















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