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Find $p, q, r$ prime numbers $p < q < r$ such that there exists natural number $n$ such that $r^2 - q^2 - p^2 = n^2$.

What I have done so far is obtaining the smallest number $p$ which is $2$. The reason is that if we assume that $p$ is odd, as $p$ is the smallest of the 3 prime numbers, then $q, r$ are also odd and thus so is $n$.

Then, if we rewrite the initial relation as: $(r - p)(r + p) = n^2 + q^2$, we see that 4 divides $(r - p)(r + p)$, so 4 must also divide $n^2 + q^2$. But as $n$ and $q$ are odd, we have that $n^2 + q^2 = (2k + 1)^2 + (2l + 1)^2 = 4(k^2 + l^2 + k + l) + 2$ which is obviously not a multiple of 4.

Therefore, $p$ must be equal to 2.

For $q$ and $r$ I tried using other similar divisors arguments (as I believe the only solution is $(2,3,7)$), but none of them work, so if you have any suggestions, I would be very thankful.

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Assume $q$ and $r$ are primes bigger than $3$. Then they are not a multiple of $3$, so their squares are $1$ mod $3$. Then $r^2-q^2-p^2$ will be $2$ mod $3$, so not a square.

So we need $q=3$. Then we need $r^2-13=n^2$ and there are only finitely many pairs of squares differing by $13$, and as you found $7^2-13=6^2$ is the only solution.

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  • $\begingroup$ Thank you a lot! This was very helpful :) $\endgroup$ – Sandel Dec 21 '18 at 18:59
  • $\begingroup$ You got quite far by yourself! Even just the observation that $(2,3,7)$ was the only solution you could find was quite useful. $\endgroup$ – SmileyCraft Dec 21 '18 at 19:07
  • $\begingroup$ $q=3$, $p=4$, $3^2+4^2=5^2$, $13^2-5^2=12^2$, so $r=13$ and $n=12$. $\endgroup$ – sirous Dec 27 '18 at 14:50
  • $\begingroup$ $4$ is not a prime. $\endgroup$ – SmileyCraft Dec 27 '18 at 15:33

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