1969 Canadian MO Problems/Problem 7: Difference between revisions
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== Problem == | == Problem == | ||
Show that there are no integers <math> | Show that there are no integers <math>a,b,c</math> for which <math>a^2+b^2-8c=6</math>. | ||
== Solution == | == Solution == | ||
Note that all [[perfect square]]s are equivalent to <math> | Note that all [[perfect square]]s are equivalent to <math>0,1,4\pmod8.</math> Hence, we have <math>a^2+b^2\equiv 6\pmod8.</math> It's impossible to obtain a sum of <math>6</math> with two of <math>0,1,4,</math> so our proof is complete. | ||
== References == | == References == | ||
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* [[1969 Canadian MO Problems|Back to Exam]] | * [[1969 Canadian MO Problems|Back to Exam]] | ||
{{Old CanadaMO box|num-b=6|num-a=8|year=1969}} | |||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
Revision as of 21:41, 17 November 2007
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.
References
| 1969 Canadian MO (Problems) | ||
| Preceded by Problem 6 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 8 |
