How do you show that, for any integer, there is a triangle with side rational lengths and that integer area?












6















In Cohen's book, Cohen's Number Theory Volume 1, the first exercise is to show that, for any integer, there is a triangle with side rational lengths such that the triangle has that integer as an area.




For example,



What are the side rational lengths for an area 2 triangle?



Given Heron's formula for a triangle of area $2$,



$$sqrt{frac{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}{16}}=2$$



How do we find the sides for a side rational triangle $(a,b,c)$ that satisfies this equation?



Another example, (9,10,17)/6 has area 1, and so on for each integer.




Looking for the method for solving the exercise, not necessarily a compendium of known triples with integer areas.











share|cite|improve this question
























  • Do you mean a single rational triangle or all of them? The simplest instance is sides $,(5,29,30)/6.$
    – Somos
    Dec 24 at 0:13












  • I just mean a single rational triangle with area $2$. Since $(9,10,17)/6$ has area $1$, and you just gave us one for area $2$, then the next question is exactly "how" did you get your simplest instance? Then, we need to find area $3$, I guess. In Cohen's Number Theory book in chapter 1, the first exercise is to show that for all integer areas, one can prove there is a rational triangle. So, "how" do we do this? We don't need all of them for the proof, just one per integer area, for each integer.
    – Pythagorus
    Dec 24 at 1:53
















6















In Cohen's book, Cohen's Number Theory Volume 1, the first exercise is to show that, for any integer, there is a triangle with side rational lengths such that the triangle has that integer as an area.




For example,



What are the side rational lengths for an area 2 triangle?



Given Heron's formula for a triangle of area $2$,



$$sqrt{frac{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}{16}}=2$$



How do we find the sides for a side rational triangle $(a,b,c)$ that satisfies this equation?



Another example, (9,10,17)/6 has area 1, and so on for each integer.




Looking for the method for solving the exercise, not necessarily a compendium of known triples with integer areas.











share|cite|improve this question
























  • Do you mean a single rational triangle or all of them? The simplest instance is sides $,(5,29,30)/6.$
    – Somos
    Dec 24 at 0:13












  • I just mean a single rational triangle with area $2$. Since $(9,10,17)/6$ has area $1$, and you just gave us one for area $2$, then the next question is exactly "how" did you get your simplest instance? Then, we need to find area $3$, I guess. In Cohen's Number Theory book in chapter 1, the first exercise is to show that for all integer areas, one can prove there is a rational triangle. So, "how" do we do this? We don't need all of them for the proof, just one per integer area, for each integer.
    – Pythagorus
    Dec 24 at 1:53














6












6








6


1






In Cohen's book, Cohen's Number Theory Volume 1, the first exercise is to show that, for any integer, there is a triangle with side rational lengths such that the triangle has that integer as an area.




For example,



What are the side rational lengths for an area 2 triangle?



Given Heron's formula for a triangle of area $2$,



$$sqrt{frac{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}{16}}=2$$



How do we find the sides for a side rational triangle $(a,b,c)$ that satisfies this equation?



Another example, (9,10,17)/6 has area 1, and so on for each integer.




Looking for the method for solving the exercise, not necessarily a compendium of known triples with integer areas.











share|cite|improve this question
















In Cohen's book, Cohen's Number Theory Volume 1, the first exercise is to show that, for any integer, there is a triangle with side rational lengths such that the triangle has that integer as an area.




For example,



What are the side rational lengths for an area 2 triangle?



Given Heron's formula for a triangle of area $2$,



$$sqrt{frac{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}{16}}=2$$



How do we find the sides for a side rational triangle $(a,b,c)$ that satisfies this equation?



Another example, (9,10,17)/6 has area 1, and so on for each integer.




Looking for the method for solving the exercise, not necessarily a compendium of known triples with integer areas.








geometry number-theory triangle diophantine-equations






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share|cite|improve this question













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edited 2 days ago









Blue

47.6k870151




47.6k870151










asked Dec 23 at 23:02









Pythagorus

1448




1448












  • Do you mean a single rational triangle or all of them? The simplest instance is sides $,(5,29,30)/6.$
    – Somos
    Dec 24 at 0:13












  • I just mean a single rational triangle with area $2$. Since $(9,10,17)/6$ has area $1$, and you just gave us one for area $2$, then the next question is exactly "how" did you get your simplest instance? Then, we need to find area $3$, I guess. In Cohen's Number Theory book in chapter 1, the first exercise is to show that for all integer areas, one can prove there is a rational triangle. So, "how" do we do this? We don't need all of them for the proof, just one per integer area, for each integer.
    – Pythagorus
    Dec 24 at 1:53


















  • Do you mean a single rational triangle or all of them? The simplest instance is sides $,(5,29,30)/6.$
    – Somos
    Dec 24 at 0:13












  • I just mean a single rational triangle with area $2$. Since $(9,10,17)/6$ has area $1$, and you just gave us one for area $2$, then the next question is exactly "how" did you get your simplest instance? Then, we need to find area $3$, I guess. In Cohen's Number Theory book in chapter 1, the first exercise is to show that for all integer areas, one can prove there is a rational triangle. So, "how" do we do this? We don't need all of them for the proof, just one per integer area, for each integer.
    – Pythagorus
    Dec 24 at 1:53
















Do you mean a single rational triangle or all of them? The simplest instance is sides $,(5,29,30)/6.$
– Somos
Dec 24 at 0:13






Do you mean a single rational triangle or all of them? The simplest instance is sides $,(5,29,30)/6.$
– Somos
Dec 24 at 0:13














I just mean a single rational triangle with area $2$. Since $(9,10,17)/6$ has area $1$, and you just gave us one for area $2$, then the next question is exactly "how" did you get your simplest instance? Then, we need to find area $3$, I guess. In Cohen's Number Theory book in chapter 1, the first exercise is to show that for all integer areas, one can prove there is a rational triangle. So, "how" do we do this? We don't need all of them for the proof, just one per integer area, for each integer.
– Pythagorus
Dec 24 at 1:53




I just mean a single rational triangle with area $2$. Since $(9,10,17)/6$ has area $1$, and you just gave us one for area $2$, then the next question is exactly "how" did you get your simplest instance? Then, we need to find area $3$, I guess. In Cohen's Number Theory book in chapter 1, the first exercise is to show that for all integer areas, one can prove there is a rational triangle. So, "how" do we do this? We don't need all of them for the proof, just one per integer area, for each integer.
– Pythagorus
Dec 24 at 1:53










3 Answers
3






active

oldest

votes


















0














Well, the formula itself Geronova triangle.



$$S_g=sqrt{(a+b+c)(a+b-c)(a-b+c)(b+c-a)}$$



If: $p,s,k,t$ -integers asked us. Then the solutions are.



$$a=(pt+ks)(k^2+t^2)ps$$



$$b=(pt-ks)((k^2+t^2)ps+(p^2+s^2)kt)$$



$$c=(pt+ks)(p^2+s^2)kt$$



$$S_g=4pskt(p^2t^2-k^2s^2)((k^2+t^2)ps+(p^2+s^2)kt)$$






share|cite|improve this answer





















  • If $p,s,k,t$ have to be integers, then the parameterization you gave does not even recover the integer areas $1$ or $2$. So, it does not help.
    – Pythagorus
    2 days ago



















0














For (p,s,k,t)=(2,1,1,1), the formula,



given by "Individ" gives us the
triangle $(a,b,c)=(12,9,15)$ and area $(A) =54$



Where $S_g=4A$



The area $A=54$ is an integer. So integer $(54)$ is represented as a triangle by the formula given by "Individ"



There are numerous formula's to represent the area's of different triangle's.



But if there is a general solution (regarding triangle's) representing all the integer's is anybody' guess.






share|cite|improve this answer








New contributor




Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
    – Leucippus
    2 days ago



















0















Partial Solution which may help .




Given $Min mathbb{N}$ we are supposed to find $a,b$ and $c in mathbb{Q}$ such that $M^2=s(s-a)(s-b)(s-c) $ where $s=frac{a+b+c}{2}$



This is equivalent to finding $a,b$ and $c$ such that $$16M^2=(a+b+c)(a-b+c)(a+b-c)(-a+b+c)$$



Let $(a-b+c)=X,(a+b-c)=Y$ and $(-a+b+c)=Z$



Then we are supposed determine $X,Y$ and $Zin mathbb{Q}^+$ such that $$16M^2=(X+Y+Z)cdot X cdot Ycdot Z.$$



Let $(X+Y+Z)=2^kcdot M$ and $X cdot Ycdot Z=frac{16M}{2^k}$



The solution to this system exists when $P^2 geq 4Q$ where $P=2^k M-X$ and $Q=frac{16M}{2^k}$



That is solution will exists for some bigger $k$ as LHS of $P^2 geq 4Q$(After making signs of all terms positive by transposing) includes $2^{2k}$ but RHS have highest exponent $2^{k+1}$



We are left to find $Y,Z in mathbb{Q}^+$ such that $(2^kM-Y-Z)(YZ)=16M{2^k}$ and $2^k M-Y-Z >0$.
Note that this is a curve in $mathbb{R}^2. Which is connected



When are rational then $Y,Z$ then $X=2^k M-Y-Z$ which is rational and hence the system of equations $(a-b+c)=X,(a+b-c)=Y$ and $(-a+b+c)=Z$ admits rational solution because they are linear equations.



Note-Other properties of Triangle are automatically satisfied because if $X>0$ then $a+c>b$ and so on.






share|cite|improve this answer























  • $P^2ge4Q$ is not necessary to conclude that if $Y$ and $Z$ are rational, then $X$ must be rational to produce integer $M$. And concluding this much is trivial toward trying to establish that integer $4M=sqrt{(X + Y + Z) cdot X cdot Y cdot Z}$ where $X$, $Y$, and $Z$ (all three) are rational, which is the question.
    – Pythagorus
    16 hours ago













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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














Well, the formula itself Geronova triangle.



$$S_g=sqrt{(a+b+c)(a+b-c)(a-b+c)(b+c-a)}$$



If: $p,s,k,t$ -integers asked us. Then the solutions are.



$$a=(pt+ks)(k^2+t^2)ps$$



$$b=(pt-ks)((k^2+t^2)ps+(p^2+s^2)kt)$$



$$c=(pt+ks)(p^2+s^2)kt$$



$$S_g=4pskt(p^2t^2-k^2s^2)((k^2+t^2)ps+(p^2+s^2)kt)$$






share|cite|improve this answer





















  • If $p,s,k,t$ have to be integers, then the parameterization you gave does not even recover the integer areas $1$ or $2$. So, it does not help.
    – Pythagorus
    2 days ago
















0














Well, the formula itself Geronova triangle.



$$S_g=sqrt{(a+b+c)(a+b-c)(a-b+c)(b+c-a)}$$



If: $p,s,k,t$ -integers asked us. Then the solutions are.



$$a=(pt+ks)(k^2+t^2)ps$$



$$b=(pt-ks)((k^2+t^2)ps+(p^2+s^2)kt)$$



$$c=(pt+ks)(p^2+s^2)kt$$



$$S_g=4pskt(p^2t^2-k^2s^2)((k^2+t^2)ps+(p^2+s^2)kt)$$






share|cite|improve this answer





















  • If $p,s,k,t$ have to be integers, then the parameterization you gave does not even recover the integer areas $1$ or $2$. So, it does not help.
    – Pythagorus
    2 days ago














0












0








0






Well, the formula itself Geronova triangle.



$$S_g=sqrt{(a+b+c)(a+b-c)(a-b+c)(b+c-a)}$$



If: $p,s,k,t$ -integers asked us. Then the solutions are.



$$a=(pt+ks)(k^2+t^2)ps$$



$$b=(pt-ks)((k^2+t^2)ps+(p^2+s^2)kt)$$



$$c=(pt+ks)(p^2+s^2)kt$$



$$S_g=4pskt(p^2t^2-k^2s^2)((k^2+t^2)ps+(p^2+s^2)kt)$$






share|cite|improve this answer












Well, the formula itself Geronova triangle.



$$S_g=sqrt{(a+b+c)(a+b-c)(a-b+c)(b+c-a)}$$



If: $p,s,k,t$ -integers asked us. Then the solutions are.



$$a=(pt+ks)(k^2+t^2)ps$$



$$b=(pt-ks)((k^2+t^2)ps+(p^2+s^2)kt)$$



$$c=(pt+ks)(p^2+s^2)kt$$



$$S_g=4pskt(p^2t^2-k^2s^2)((k^2+t^2)ps+(p^2+s^2)kt)$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 2 days ago









individ

3,2421816




3,2421816












  • If $p,s,k,t$ have to be integers, then the parameterization you gave does not even recover the integer areas $1$ or $2$. So, it does not help.
    – Pythagorus
    2 days ago


















  • If $p,s,k,t$ have to be integers, then the parameterization you gave does not even recover the integer areas $1$ or $2$. So, it does not help.
    – Pythagorus
    2 days ago
















If $p,s,k,t$ have to be integers, then the parameterization you gave does not even recover the integer areas $1$ or $2$. So, it does not help.
– Pythagorus
2 days ago




If $p,s,k,t$ have to be integers, then the parameterization you gave does not even recover the integer areas $1$ or $2$. So, it does not help.
– Pythagorus
2 days ago











0














For (p,s,k,t)=(2,1,1,1), the formula,



given by "Individ" gives us the
triangle $(a,b,c)=(12,9,15)$ and area $(A) =54$



Where $S_g=4A$



The area $A=54$ is an integer. So integer $(54)$ is represented as a triangle by the formula given by "Individ"



There are numerous formula's to represent the area's of different triangle's.



But if there is a general solution (regarding triangle's) representing all the integer's is anybody' guess.






share|cite|improve this answer








New contributor




Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
    – Leucippus
    2 days ago
















0














For (p,s,k,t)=(2,1,1,1), the formula,



given by "Individ" gives us the
triangle $(a,b,c)=(12,9,15)$ and area $(A) =54$



Where $S_g=4A$



The area $A=54$ is an integer. So integer $(54)$ is represented as a triangle by the formula given by "Individ"



There are numerous formula's to represent the area's of different triangle's.



But if there is a general solution (regarding triangle's) representing all the integer's is anybody' guess.






share|cite|improve this answer








New contributor




Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
    – Leucippus
    2 days ago














0












0








0






For (p,s,k,t)=(2,1,1,1), the formula,



given by "Individ" gives us the
triangle $(a,b,c)=(12,9,15)$ and area $(A) =54$



Where $S_g=4A$



The area $A=54$ is an integer. So integer $(54)$ is represented as a triangle by the formula given by "Individ"



There are numerous formula's to represent the area's of different triangle's.



But if there is a general solution (regarding triangle's) representing all the integer's is anybody' guess.






share|cite|improve this answer








New contributor




Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









For (p,s,k,t)=(2,1,1,1), the formula,



given by "Individ" gives us the
triangle $(a,b,c)=(12,9,15)$ and area $(A) =54$



Where $S_g=4A$



The area $A=54$ is an integer. So integer $(54)$ is represented as a triangle by the formula given by "Individ"



There are numerous formula's to represent the area's of different triangle's.



But if there is a general solution (regarding triangle's) representing all the integer's is anybody' guess.







share|cite|improve this answer








New contributor




Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this answer



share|cite|improve this answer






New contributor




Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









answered 2 days ago









Sam

1




1




New contributor




Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
    – Leucippus
    2 days ago


















  • This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
    – Leucippus
    2 days ago
















This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
– Leucippus
2 days ago




This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
– Leucippus
2 days ago











0















Partial Solution which may help .




Given $Min mathbb{N}$ we are supposed to find $a,b$ and $c in mathbb{Q}$ such that $M^2=s(s-a)(s-b)(s-c) $ where $s=frac{a+b+c}{2}$



This is equivalent to finding $a,b$ and $c$ such that $$16M^2=(a+b+c)(a-b+c)(a+b-c)(-a+b+c)$$



Let $(a-b+c)=X,(a+b-c)=Y$ and $(-a+b+c)=Z$



Then we are supposed determine $X,Y$ and $Zin mathbb{Q}^+$ such that $$16M^2=(X+Y+Z)cdot X cdot Ycdot Z.$$



Let $(X+Y+Z)=2^kcdot M$ and $X cdot Ycdot Z=frac{16M}{2^k}$



The solution to this system exists when $P^2 geq 4Q$ where $P=2^k M-X$ and $Q=frac{16M}{2^k}$



That is solution will exists for some bigger $k$ as LHS of $P^2 geq 4Q$(After making signs of all terms positive by transposing) includes $2^{2k}$ but RHS have highest exponent $2^{k+1}$



We are left to find $Y,Z in mathbb{Q}^+$ such that $(2^kM-Y-Z)(YZ)=16M{2^k}$ and $2^k M-Y-Z >0$.
Note that this is a curve in $mathbb{R}^2. Which is connected



When are rational then $Y,Z$ then $X=2^k M-Y-Z$ which is rational and hence the system of equations $(a-b+c)=X,(a+b-c)=Y$ and $(-a+b+c)=Z$ admits rational solution because they are linear equations.



Note-Other properties of Triangle are automatically satisfied because if $X>0$ then $a+c>b$ and so on.






share|cite|improve this answer























  • $P^2ge4Q$ is not necessary to conclude that if $Y$ and $Z$ are rational, then $X$ must be rational to produce integer $M$. And concluding this much is trivial toward trying to establish that integer $4M=sqrt{(X + Y + Z) cdot X cdot Y cdot Z}$ where $X$, $Y$, and $Z$ (all three) are rational, which is the question.
    – Pythagorus
    16 hours ago


















0















Partial Solution which may help .




Given $Min mathbb{N}$ we are supposed to find $a,b$ and $c in mathbb{Q}$ such that $M^2=s(s-a)(s-b)(s-c) $ where $s=frac{a+b+c}{2}$



This is equivalent to finding $a,b$ and $c$ such that $$16M^2=(a+b+c)(a-b+c)(a+b-c)(-a+b+c)$$



Let $(a-b+c)=X,(a+b-c)=Y$ and $(-a+b+c)=Z$



Then we are supposed determine $X,Y$ and $Zin mathbb{Q}^+$ such that $$16M^2=(X+Y+Z)cdot X cdot Ycdot Z.$$



Let $(X+Y+Z)=2^kcdot M$ and $X cdot Ycdot Z=frac{16M}{2^k}$



The solution to this system exists when $P^2 geq 4Q$ where $P=2^k M-X$ and $Q=frac{16M}{2^k}$



That is solution will exists for some bigger $k$ as LHS of $P^2 geq 4Q$(After making signs of all terms positive by transposing) includes $2^{2k}$ but RHS have highest exponent $2^{k+1}$



We are left to find $Y,Z in mathbb{Q}^+$ such that $(2^kM-Y-Z)(YZ)=16M{2^k}$ and $2^k M-Y-Z >0$.
Note that this is a curve in $mathbb{R}^2. Which is connected



When are rational then $Y,Z$ then $X=2^k M-Y-Z$ which is rational and hence the system of equations $(a-b+c)=X,(a+b-c)=Y$ and $(-a+b+c)=Z$ admits rational solution because they are linear equations.



Note-Other properties of Triangle are automatically satisfied because if $X>0$ then $a+c>b$ and so on.






share|cite|improve this answer























  • $P^2ge4Q$ is not necessary to conclude that if $Y$ and $Z$ are rational, then $X$ must be rational to produce integer $M$. And concluding this much is trivial toward trying to establish that integer $4M=sqrt{(X + Y + Z) cdot X cdot Y cdot Z}$ where $X$, $Y$, and $Z$ (all three) are rational, which is the question.
    – Pythagorus
    16 hours ago
















0












0








0







Partial Solution which may help .




Given $Min mathbb{N}$ we are supposed to find $a,b$ and $c in mathbb{Q}$ such that $M^2=s(s-a)(s-b)(s-c) $ where $s=frac{a+b+c}{2}$



This is equivalent to finding $a,b$ and $c$ such that $$16M^2=(a+b+c)(a-b+c)(a+b-c)(-a+b+c)$$



Let $(a-b+c)=X,(a+b-c)=Y$ and $(-a+b+c)=Z$



Then we are supposed determine $X,Y$ and $Zin mathbb{Q}^+$ such that $$16M^2=(X+Y+Z)cdot X cdot Ycdot Z.$$



Let $(X+Y+Z)=2^kcdot M$ and $X cdot Ycdot Z=frac{16M}{2^k}$



The solution to this system exists when $P^2 geq 4Q$ where $P=2^k M-X$ and $Q=frac{16M}{2^k}$



That is solution will exists for some bigger $k$ as LHS of $P^2 geq 4Q$(After making signs of all terms positive by transposing) includes $2^{2k}$ but RHS have highest exponent $2^{k+1}$



We are left to find $Y,Z in mathbb{Q}^+$ such that $(2^kM-Y-Z)(YZ)=16M{2^k}$ and $2^k M-Y-Z >0$.
Note that this is a curve in $mathbb{R}^2. Which is connected



When are rational then $Y,Z$ then $X=2^k M-Y-Z$ which is rational and hence the system of equations $(a-b+c)=X,(a+b-c)=Y$ and $(-a+b+c)=Z$ admits rational solution because they are linear equations.



Note-Other properties of Triangle are automatically satisfied because if $X>0$ then $a+c>b$ and so on.






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Partial Solution which may help .




Given $Min mathbb{N}$ we are supposed to find $a,b$ and $c in mathbb{Q}$ such that $M^2=s(s-a)(s-b)(s-c) $ where $s=frac{a+b+c}{2}$



This is equivalent to finding $a,b$ and $c$ such that $$16M^2=(a+b+c)(a-b+c)(a+b-c)(-a+b+c)$$



Let $(a-b+c)=X,(a+b-c)=Y$ and $(-a+b+c)=Z$



Then we are supposed determine $X,Y$ and $Zin mathbb{Q}^+$ such that $$16M^2=(X+Y+Z)cdot X cdot Ycdot Z.$$



Let $(X+Y+Z)=2^kcdot M$ and $X cdot Ycdot Z=frac{16M}{2^k}$



The solution to this system exists when $P^2 geq 4Q$ where $P=2^k M-X$ and $Q=frac{16M}{2^k}$



That is solution will exists for some bigger $k$ as LHS of $P^2 geq 4Q$(After making signs of all terms positive by transposing) includes $2^{2k}$ but RHS have highest exponent $2^{k+1}$



We are left to find $Y,Z in mathbb{Q}^+$ such that $(2^kM-Y-Z)(YZ)=16M{2^k}$ and $2^k M-Y-Z >0$.
Note that this is a curve in $mathbb{R}^2. Which is connected



When are rational then $Y,Z$ then $X=2^k M-Y-Z$ which is rational and hence the system of equations $(a-b+c)=X,(a+b-c)=Y$ and $(-a+b+c)=Z$ admits rational solution because they are linear equations.



Note-Other properties of Triangle are automatically satisfied because if $X>0$ then $a+c>b$ and so on.







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share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered 2 days ago









Rakesh Bhatt

851114




851114












  • $P^2ge4Q$ is not necessary to conclude that if $Y$ and $Z$ are rational, then $X$ must be rational to produce integer $M$. And concluding this much is trivial toward trying to establish that integer $4M=sqrt{(X + Y + Z) cdot X cdot Y cdot Z}$ where $X$, $Y$, and $Z$ (all three) are rational, which is the question.
    – Pythagorus
    16 hours ago




















  • $P^2ge4Q$ is not necessary to conclude that if $Y$ and $Z$ are rational, then $X$ must be rational to produce integer $M$. And concluding this much is trivial toward trying to establish that integer $4M=sqrt{(X + Y + Z) cdot X cdot Y cdot Z}$ where $X$, $Y$, and $Z$ (all three) are rational, which is the question.
    – Pythagorus
    16 hours ago


















$P^2ge4Q$ is not necessary to conclude that if $Y$ and $Z$ are rational, then $X$ must be rational to produce integer $M$. And concluding this much is trivial toward trying to establish that integer $4M=sqrt{(X + Y + Z) cdot X cdot Y cdot Z}$ where $X$, $Y$, and $Z$ (all three) are rational, which is the question.
– Pythagorus
16 hours ago






$P^2ge4Q$ is not necessary to conclude that if $Y$ and $Z$ are rational, then $X$ must be rational to produce integer $M$. And concluding this much is trivial toward trying to establish that integer $4M=sqrt{(X + Y + Z) cdot X cdot Y cdot Z}$ where $X$, $Y$, and $Z$ (all three) are rational, which is the question.
– Pythagorus
16 hours ago




















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