Sum and Product of Root












-1












$begingroup$


Suppose the roots of the polynomial $x^2+mx+n$ are positive prime integers (not necessarily distinct). Given that $m<20$ how many possible values of $n$ are there?










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  • $begingroup$
    What have you tried? Have you, say, found some possible values for $n$?
    $endgroup$
    – lulu
    Jan 18 at 2:12






  • 3




    $begingroup$
    Also, did you mean to write $|m|<20$? After all, if the roots are both positive then $m$ is negative, so, as written, there are infinitely many possible values of $n$.
    $endgroup$
    – lulu
    Jan 18 at 2:16
















-1












$begingroup$


Suppose the roots of the polynomial $x^2+mx+n$ are positive prime integers (not necessarily distinct). Given that $m<20$ how many possible values of $n$ are there?










share|cite|improve this question











$endgroup$












  • $begingroup$
    What have you tried? Have you, say, found some possible values for $n$?
    $endgroup$
    – lulu
    Jan 18 at 2:12






  • 3




    $begingroup$
    Also, did you mean to write $|m|<20$? After all, if the roots are both positive then $m$ is negative, so, as written, there are infinitely many possible values of $n$.
    $endgroup$
    – lulu
    Jan 18 at 2:16














-1












-1








-1


0



$begingroup$


Suppose the roots of the polynomial $x^2+mx+n$ are positive prime integers (not necessarily distinct). Given that $m<20$ how many possible values of $n$ are there?










share|cite|improve this question











$endgroup$




Suppose the roots of the polynomial $x^2+mx+n$ are positive prime integers (not necessarily distinct). Given that $m<20$ how many possible values of $n$ are there?







algebra-precalculus elementary-number-theory






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share|cite|improve this question













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share|cite|improve this question








edited Jan 18 at 2:10









clathratus

5,1441439




5,1441439










asked Jan 18 at 2:09









sumisumi

654




654












  • $begingroup$
    What have you tried? Have you, say, found some possible values for $n$?
    $endgroup$
    – lulu
    Jan 18 at 2:12






  • 3




    $begingroup$
    Also, did you mean to write $|m|<20$? After all, if the roots are both positive then $m$ is negative, so, as written, there are infinitely many possible values of $n$.
    $endgroup$
    – lulu
    Jan 18 at 2:16


















  • $begingroup$
    What have you tried? Have you, say, found some possible values for $n$?
    $endgroup$
    – lulu
    Jan 18 at 2:12






  • 3




    $begingroup$
    Also, did you mean to write $|m|<20$? After all, if the roots are both positive then $m$ is negative, so, as written, there are infinitely many possible values of $n$.
    $endgroup$
    – lulu
    Jan 18 at 2:16
















$begingroup$
What have you tried? Have you, say, found some possible values for $n$?
$endgroup$
– lulu
Jan 18 at 2:12




$begingroup$
What have you tried? Have you, say, found some possible values for $n$?
$endgroup$
– lulu
Jan 18 at 2:12




3




3




$begingroup$
Also, did you mean to write $|m|<20$? After all, if the roots are both positive then $m$ is negative, so, as written, there are infinitely many possible values of $n$.
$endgroup$
– lulu
Jan 18 at 2:16




$begingroup$
Also, did you mean to write $|m|<20$? After all, if the roots are both positive then $m$ is negative, so, as written, there are infinitely many possible values of $n$.
$endgroup$
– lulu
Jan 18 at 2:16










1 Answer
1






active

oldest

votes


















1












$begingroup$

Assuming $m < 20$ means $|m| < 20$,



Say the roots are $p_1$ and $p_2$ (needn't be distinct)



then, the quadratic becomes $x^2 - (p_1 + p_2)x + p_1p_2$



$m = - (p_1 + p_2)$ and $n = p_1p_2$



So, $p_1$ and $p_2$ can take all combinations (repeat) from$ ({2, 3, 5, 7, 11, 13, 17} | p_1 + p_2 < 20)$



number of $n = 18$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You should have $17$ instead of $19$, and unless I'm missing something, there are only $18$ valid combinations.
    $endgroup$
    – Daniel Mathias
    Jan 18 at 3:23










  • $begingroup$
    @DanielMathias Yeah, I've repeated some. I corrected it :)
    $endgroup$
    – Neil
    Jan 18 at 3:54












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1 Answer
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active

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1 Answer
1






active

oldest

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active

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active

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1












$begingroup$

Assuming $m < 20$ means $|m| < 20$,



Say the roots are $p_1$ and $p_2$ (needn't be distinct)



then, the quadratic becomes $x^2 - (p_1 + p_2)x + p_1p_2$



$m = - (p_1 + p_2)$ and $n = p_1p_2$



So, $p_1$ and $p_2$ can take all combinations (repeat) from$ ({2, 3, 5, 7, 11, 13, 17} | p_1 + p_2 < 20)$



number of $n = 18$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You should have $17$ instead of $19$, and unless I'm missing something, there are only $18$ valid combinations.
    $endgroup$
    – Daniel Mathias
    Jan 18 at 3:23










  • $begingroup$
    @DanielMathias Yeah, I've repeated some. I corrected it :)
    $endgroup$
    – Neil
    Jan 18 at 3:54
















1












$begingroup$

Assuming $m < 20$ means $|m| < 20$,



Say the roots are $p_1$ and $p_2$ (needn't be distinct)



then, the quadratic becomes $x^2 - (p_1 + p_2)x + p_1p_2$



$m = - (p_1 + p_2)$ and $n = p_1p_2$



So, $p_1$ and $p_2$ can take all combinations (repeat) from$ ({2, 3, 5, 7, 11, 13, 17} | p_1 + p_2 < 20)$



number of $n = 18$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You should have $17$ instead of $19$, and unless I'm missing something, there are only $18$ valid combinations.
    $endgroup$
    – Daniel Mathias
    Jan 18 at 3:23










  • $begingroup$
    @DanielMathias Yeah, I've repeated some. I corrected it :)
    $endgroup$
    – Neil
    Jan 18 at 3:54














1












1








1





$begingroup$

Assuming $m < 20$ means $|m| < 20$,



Say the roots are $p_1$ and $p_2$ (needn't be distinct)



then, the quadratic becomes $x^2 - (p_1 + p_2)x + p_1p_2$



$m = - (p_1 + p_2)$ and $n = p_1p_2$



So, $p_1$ and $p_2$ can take all combinations (repeat) from$ ({2, 3, 5, 7, 11, 13, 17} | p_1 + p_2 < 20)$



number of $n = 18$






share|cite|improve this answer











$endgroup$



Assuming $m < 20$ means $|m| < 20$,



Say the roots are $p_1$ and $p_2$ (needn't be distinct)



then, the quadratic becomes $x^2 - (p_1 + p_2)x + p_1p_2$



$m = - (p_1 + p_2)$ and $n = p_1p_2$



So, $p_1$ and $p_2$ can take all combinations (repeat) from$ ({2, 3, 5, 7, 11, 13, 17} | p_1 + p_2 < 20)$



number of $n = 18$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 18 at 3:52

























answered Jan 18 at 3:11









NeilNeil

113




113












  • $begingroup$
    You should have $17$ instead of $19$, and unless I'm missing something, there are only $18$ valid combinations.
    $endgroup$
    – Daniel Mathias
    Jan 18 at 3:23










  • $begingroup$
    @DanielMathias Yeah, I've repeated some. I corrected it :)
    $endgroup$
    – Neil
    Jan 18 at 3:54


















  • $begingroup$
    You should have $17$ instead of $19$, and unless I'm missing something, there are only $18$ valid combinations.
    $endgroup$
    – Daniel Mathias
    Jan 18 at 3:23










  • $begingroup$
    @DanielMathias Yeah, I've repeated some. I corrected it :)
    $endgroup$
    – Neil
    Jan 18 at 3:54
















$begingroup$
You should have $17$ instead of $19$, and unless I'm missing something, there are only $18$ valid combinations.
$endgroup$
– Daniel Mathias
Jan 18 at 3:23




$begingroup$
You should have $17$ instead of $19$, and unless I'm missing something, there are only $18$ valid combinations.
$endgroup$
– Daniel Mathias
Jan 18 at 3:23












$begingroup$
@DanielMathias Yeah, I've repeated some. I corrected it :)
$endgroup$
– Neil
Jan 18 at 3:54




$begingroup$
@DanielMathias Yeah, I've repeated some. I corrected it :)
$endgroup$
– Neil
Jan 18 at 3:54


















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