Find $p, q, r$ prime numbers such that $r^2 - q^2 - p^2 = n^2$.

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.

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