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

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

$$n^2+182n=k^2$$

$$(k-n)(k+n)=182n$$

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?

• 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).
– lulu
Oct 18, 2020 at 15:04
• Do you want $A$ an integer or just a rational number? Oct 18, 2020 at 15:18
• @lulu thank you very much, great hint
– user814992
Oct 18, 2020 at 15:19
• @J.W.Tanner A is rational
– user814992
Oct 18, 2020 at 15:19
• hint: 4 solutions (only). Oct 18, 2020 at 15:27

$$n^2+182n=k^2\implies (n+91)^2=k^2+91^2\implies (n+91)^2-k^2=91^2\implies$$
$$(n+91+k)(n+91-k)=7\times7\times13\times13.$$
Can you solve $$(n+91-k,n+91+k)=(7,7\times13\times13)$$ or $$(13,7\times7\times13)$$
or $$(7\times13,7\times13)$$ or $$(7\times7,13\times13)$$?
• Note that $(n+91-k,n+91+k)=(7\times13,7\times13)$ would imply $n=k=0$, which was excluded by OP Oct 18, 2020 at 16:00