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

## Problem

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

## Solution

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

## References

 1969 Canadian MO (Problems) Preceded byProblem 6 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • Followed byProblem 8