Find the values for which positive integer $n$ makes $A=\sqrt{n(n+182)}$ a rational number

I tried to solve it in the following way:

$n(n+182)=k^2$ where k is an integer



From here I tried getting all the divisors of $182=2*91$ and try solving it using divisibility rules. However I did not succeed.

Could you please show me a clever and intuitive approach to solving the problem?

  • 3
    $\begingroup$ Hint: Note that $(n+91)^2=a^2$ is, of course, a square so if $n^2+182n=b^2$ is also a square then we must have $a^2-b^2=91^2$. (The idea here is that it is hard for two squares to be near each other, so search for a known square near the desired one). $\endgroup$
    – lulu
    Oct 18, 2020 at 15:04
  • $\begingroup$ Do you want $A$ an integer or just a rational number? $\endgroup$ Oct 18, 2020 at 15:18
  • $\begingroup$ @lulu thank you very much, great hint $\endgroup$
    – user814992
    Oct 18, 2020 at 15:19
  • $\begingroup$ @J.W.Tanner A is rational $\endgroup$
    – user814992
    Oct 18, 2020 at 15:19
  • $\begingroup$ hint: 4 solutions (only). $\endgroup$
    – Jean Marie
    Oct 18, 2020 at 15:27

1 Answer 1


$n^2+182n=k^2\implies (n+91)^2=k^2+91^2\implies (n+91)^2-k^2=91^2\implies$

$ (n+91+k)(n+91-k)=7\times7\times13\times13.$

Can you solve $(n+91-k,n+91+k)=(7,7\times13\times13)$ or $(13,7\times7\times13)$

or $(7\times13,7\times13)$ or $(7\times7,13\times13)$?

  • $\begingroup$ Note that $(n+91-k,n+91+k)=(7\times13,7\times13)$ would imply $n=k=0$, which was excluded by OP $\endgroup$ Oct 18, 2020 at 16:00

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