Mathematical Reflections 2012, Issue 1 - Problem J220

Problem:
Find the least prime $p$ for which $p=a^2_k+kb^2_k, k=1,\ldots,5$, for some $(a_k,b_k)$ in $\mathbb{Z} \times \mathbb{Z}$.

Proposed by Cosmin Pohoata.

Solution:

We have $p \equiv a^2_k \pmod{k}$ for $k=2,\ldots,5$, so $$\begin{array}{lll} p \equiv 1 \pmod{2} \\ p \equiv 1 \pmod{3} \\ p \equiv 1 \pmod{4} \\ p \equiv 1,4 \pmod{5}. \end{array}$$ Then $p=1+60n$ or $p=49+60n$ where $n \in \mathbb{N}$. Clearly $p \neq 1,49$. Moreover, $p \neq 61,109,121,169,181,229$ since $121,169$ are not primes and the equation
$$a^2_2+2b^2_2 \equiv 5 \pmod{8}$$ has no solution. We conclude that $p=241$ is the least prime which satisfies the given conditions, since
$$241=15^2+4^2=13^2+2\cdot6^2=7^2+3\cdot8^2=15^2+4\cdot2^2=14^2+5\cdot3^2.$$