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

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== References ==
 
== References ==
 
* [[1969 Canadian MO Problems/Problem 6|Previous Problem]]
 
* [[1969 Canadian MO Problems/Problem 8|Next Problem]]
 
* [[1969 Canadian MO Problems|Back to Exam]]
 
 
  
 
{{Old CanadaMO box|num-b=6|num-a=8|year=1969}}
 
{{Old CanadaMO box|num-b=6|num-a=8|year=1969}}
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]

Latest revision as of 10:22, 29 April 2008

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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