# Difference between revisions of "1969 Canadian MO Problems/Problem 7"

## Problem

Show that there are no integers $\displaystyle a,b,c$ for which $\displaystyle a^2+b^2-8c=6$.

## Solution

Note that all perfect squares are equivalent to $\displaystyle 0,1,4\pmod8.$ Hence, we have $\displaystyle a^2+b^2\equiv 6\pmod8.$ It's impossible to obtain a sum of $\displaystyle 6$ with two of $\displaystyle 0,1,4,$ so our proof is complete.