Difference between revisions of "1969 Canadian MO Problems/Problem 7"
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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>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 == | |
| − | + | ||
| + | {{Old CanadaMO box|num-b=6|num-a=8|year=1969}} | ||
| + | [[Category:Intermediate Number Theory Problems]] | ||
Latest revision as of 09:22, 29 April 2008
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 |
