# Prime numbers $p<q<r$ such that $r^2-q^2-p^2$ is a perfect square.

Find the prime numbers $$p such that $$r^2-q^2-p^2$$ is a perfect square.

I think the only solution is (2,3,7) but i cannot prove it. The equation would be $$r^2-q^2-p^2=k^2$$ equivalently $$q^2+p^2+k^2=r^2$$ which is somehow a classical diophantine equation (of the pythagorean quadruples) which have a parametrization but how do i use it? The problem is from a magazine at 8 th grade and i don't think the solution is supposed to use the general solution.

Actually the general solution of $$q^2+p^2+k^2=r^2$$ is given by: $$q=a^2+b^2-c^2-d^2$$

$$p=2(mq+np)$$

$$k=2(nq-mp)$$

$$r=a^2+b^2+c^2+d^2$$ IT would follow that one of the numbers q or p is even and prime, so it is 2.

• Ah, it might be in the Olympiad Corner – Dietrich Burde Mar 9 at 19:53
• @amarius8312 I have noticed that you haven't accepted any of the 14 questions you've made on MSE. Please consider accepting the answers you like the most ;) PS: You get 2 extra reputation points whenever you accept an answer – Dr. Mathva Mar 29 at 15:05

Hint

All prime numbers $$p\geq5$$ satisfy $$p\equiv \pm 1\mod 6$$

Therefore, if $$p,q,r\geq5$$

$$r^2-q^2-p^2\equiv (\pm1)^2-(\pm1)^2-(\pm1)^2\equiv 1-1-1\equiv5\mod 6$$

Observe now, that there's no perfect square with the residue $$5$$ modulo 6.
This follows from the simple fact that (taking the equations modulo $$6$$)

$$\begin{array}l 1^2\equiv\color{red}1\\ 2^2\equiv \color{red}4\\ 3^2\equiv9\equiv\color{red}3\\4^2\equiv16\equiv \color{red}4\\5^2\equiv25\equiv \color{red}1 \\6^2\equiv\color{red}0\end{array}$$

Can you end it now?

Your guess is correct: $$(2,3,7)$$ is the unique solution. There is a simple proof using two basic and easy to verify facts:

• if $$m$$ is odd, then $$m^2\equiv 1\pmod 4$$;
• if $$m$$ is not divisible by $$3$$, then $$m^2\equiv 1\pmod 3$$.

We proceed as follows.

Write $$r^2-p^2-q^2=k^2$$ with an integer $$k$$.

If we had $$p>2$$, this would imply $$p^2\equiv q^2\equiv r^2\equiv 1\pmod 4$$, leading to $$k^2\equiv r^2-p^2-q^2\equiv 3\pmod 4$$, which contradicts the first basic fact above. Thus $$p=2$$, whence $$r^2=q^2+k^2+4$$.

If we had $$q>3$$, this would imply $$r^2\equiv q^2\equiv 1\pmod 3$$ leading to $$k^2\equiv 2\pmod 3$$, which contradicts the second basic fact above. Thus $$q=3$$, and, as a result, $$r^2=13+k^2$$. Consequently, $$(r-k)(r+k)=13$$, which is only possible if $$r-k=1$$ and $$r+k=13$$. This yields $$r=7$$, $$k=6$$.

To summarize, the only solution is $$p=2$$, $$q=3$$, $$r=7$$.