For a positive integer $n$ let $H_n=\sum_{ k=1}^{ n}1/n$ be the $n$th harmonic number. In this note we prove that for any prime $p\ge 7$, $$ \sum_{ k=1}^{ p-1}\frac{ H_k}{ k^2}\equiv \sum_{ k=1}^{ p-1}\frac{ H_k^2}{ k} \equiv\frac{ 3}{ 2p}\sum_{ k=1}^{ p-1}\frac{ 1}{ k^2}\pmod{ p^2}. $$ Notice that the first part of this congruence is recently proposed by R. Tauraso as a problem in Amer. Math. Monthly. In our elementary proof of the second part of the above congruence we use certain classical congruences modulo a prime and the square of a prime, some congruences involving harmonic numbers and a combinatorial identity due to V. Hern\'{ a}ndez.
Categories: math.NT