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.
Best Answer
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.