The difference between the arithmetic mean and the geometric mean of two positive integers $\displaystyle m$ and $\displaystyle n$ is 1. Prove that $\displaystyle \frac {m}{2}$ and $\displaystyle \frac {n}{2}$ are perfect squares.
The difference between the arithmetic mean and the geometric mean of two positive integers $\displaystyle m$ and $\displaystyle n$ is 1. Prove that $\displaystyle \frac {m}{2}$ and $\displaystyle \frac {n}{2}$ are perfect squares.
$\displaystyle \frac {m + n}{2} = 1 + \sqrt {mn}$
Clearly $\displaystyle mn$ is a perfect square, therefore $\displaystyle m = a^2 t, \, n = b^2 t$ for some square free t
Therefore
$\displaystyle \frac {t(a^2 + b^2)}{2} = 1 + abt \Rightarrow t(a - b)^2 = 2 \Rightarrow t = 2$ as required