Peter wrote: If an integer $n$ is such that $7n$ is the form $a^2 +3b^2$, prove that $n$ is also of that form. In fact, let $A$ and $B$ be two integers satisfying $7n=A^2+3B^2$. Then, $7\mid A^2+3B^2$, so that also $7\mid\left(A^2+3B^2\right)-7B^2=A^2-4B^2=\left(A+2B\right)\left(A-2B\right)$. Thus, since $7$ is prime, we have $7\mid A+2B$ or $7\mid A-2B$. If $7\mid A+2B$, then $\frac{A+2B}{7}$ is an integer; then, by writing $\left(2\cdot\frac{A+2B}{7}-B\right)^2+3\left(\frac{A+2B}{7}\right)^2=\frac{A^2+3B^2}{7}=\frac{7n}{7}=n$, we obtain a representation of $n$ in the form $a^2+3b^2$. If $7\mid A-2B$, then $\frac{A-2B}{7}$ is an integer; then, by writing $\left(2\cdot\frac{A-2B}{7}+B\right)^2+3\left(\frac{A-2B}{7}\right)^2=\frac{A^2+3B^2}{7}=\frac{7n}{7}=n$, we obtain a representation of $n$ in the form $a^2+3b^2$. In both cases, we have seen that $n$ can be represented in the form $a^2+3b^2$. This completes the solution. Darij

Here's the official solution:

Attachments:
India MO 1998 - Problem 1.pdf (24kb)