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 Followed by
Problem 8
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