1969 Canadian MO Problems/Problem 7: Difference between revisions
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== Problem == | == Problem == | ||
Show that there are no integers <math>\displaystyle a,b,c</math> for which <math>\displaystyle a^2+b^2-8c=6</math>. | Show that there are no integers <math>\displaystyle a,b,c</math> for which <math>\displaystyle a^2+b^2-8c=6</math>. | ||
== Solution == | == Solution == | ||
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. | 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. | ||
== References == | |||
* [[1969 Canadian MO Problems/Problem 6|Previous Problem]] | * [[1969 Canadian MO Problems/Problem 6|Previous Problem]] | ||
* [[1969 Canadian MO Problems/Problem 8|Next Problem]] | * [[1969 Canadian MO Problems/Problem 8|Next Problem]] | ||
* [[1969 Canadian MO Problems|Back to Exam]] | * [[1969 Canadian MO Problems|Back to Exam]] | ||
[[Category:Intermediate Number Theory Problems]] | |||
Revision as of 21:53, 5 September 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.
