April 5th, 2018, 07:00 PM |
#21 | |

Math Team Joined: Dec 2013 From: Colombia Posts: 7,308 Thanks: 2443 Math Focus: Mainly analysis and algebra | Quote:
It's easy to see that, for integer solutions, you need $m$ and $n$ either both odd or both even. This isn't a problem for most numbers as for odd $k$ you may take $(m,n)=(k,1)$ (other pairs may exist); while for $k$ divisible by 4, $(m,n)=(\frac k 2,2)$ works (again other pairs may exist). Even numbers not divisible by 4 won't have solutions.
Last edited by v8archie; April 5th, 2018 at 07:03 PM.
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