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

 
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== Solution ==
 
== Solution ==
Note that all perfect squares are equivilant 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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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 13:54, 28 July 2006

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.


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