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1969 Canadian MO Problems/Problem 7

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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 by
Problem 6
1 2 3 4 5 6 7 8 9 10 Followed by
Problem 8
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