Difference between revisions of "1969 Canadian MO Problems/Problem 7"
m |
|||
| Line 1: | Line 1: | ||
== 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.
