Difference between revisions of "1969 Canadian MO Problems/Problem 7"
(box) |
(Useless links) |
||
| Line 8: | Line 8: | ||
== References == | == References == | ||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
{{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 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 |
