Saudi Arabia Mathematical Competitions

Assume that $q_1^4+q_2^4+q_3^4+q_4^4+q_5^4=2k(2k+2)$ for some positive integer $k$. That is $$q_1^4+q_2^4+q_3^4+q_4^4+q_5^4=4k(k+1)$$ If $p$ is a prime, $p\ne 2$, then $p^4\equiv 1(\mathrm{mod}8)$. Assume that $q_1\le q_2\le q_3\le q_4\le q_5$. A simple parity argument shows that $q_1=2$. If $q_2\ne 2$, then we have $$q_1^4+q_2^4+q_3^4+q_4^4+q_5^4\equiv 4(\mathrm{mod}8)$$ Since $8\mid 4k(k+1)-16$, we get a contradiction. Therefore $q_2=2$. In similar way it follows $q_3=q_4=q_5=2$. The only solution is $q_1=q_2=q_3=q_4=q_5=2$, and $k=4$.