Difference between revisions of "1969 Canadian MO Problems/Problem 7"
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== Solution == | == Solution == | ||
| − | Note that all perfect | + | Note that all [[perfect square]]s are equivalent to <math>\displaystyle 0,1,4\pmod8.</math> Hence, we have <math>\displaystyle a^2+b^2\equiv 6\pmod8.</math> It's impossible to obtain a sum of <math>\displaystyle 6</math> with two of <math>\displaystyle 0,1,4,</math> so our proof is complete. |
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Revision as of 12:54, 28 July 2006
Problem
Show that there are no integers 
for which 
.
Solution
Note that all perfect squares are equivalent to 
Hence, we have 
It's impossible to obtain a sum of 
with two of 
so our proof is complete.
