Number Theory Question

[AvgStudent]

I'm trying to figure out whether this question can be answered without guessing and checking.

If [imath]m^2+n^2=40[/imath], where [imath]m<0<n;m,n \in \Z[/imath]. What's the value of [imath](m+n)^2?[/imath]

My attempt:
By primality test, [imath]m,n < \sqrt{40} \approx 6[/imath], so the factors are [imath] |m|,|n| \in \{1,2,...6\}[/imath]. Squaring them [imath]1,4,...36[/imath]. It's obvious that [imath]4+36=40[/imath].
So [imath](m+n)^2=(-2+6)^2=16.[/imath] or [imath](-6+2)^2=16[/imath]

 

[topsquark]

I'm trying to figure out whether this question can be answered without guessing and checking.

If [imath]m^2+n^2=40[/imath], where [imath]m<0<n;m,n \in \Z[/imath]. What's the value of [imath](m+n)^2?[/imath]

My attempt:
By primality test, [imath]m,n < \sqrt{40} \approx 6[/imath], so the factors are [imath] |m|,|n| \in \{1,2,...6\}[/imath]. Squaring them [imath]1,4,...36[/imath]. It's obvious that [imath]4+36=40[/imath].
So [imath](m+n)^2=(-2+6)^2=16.[/imath] or [imath](-6+2)^2=16[/imath]

I'm not certain but I don't think that there is any other way. Diophantine equations are like that. Usually the method is to restrict the possible solution set as much as possible and if you have a set left over, you check them.

-Dan

 

[BigBeachBanana]

Usually the method is to restrict the possible solution set as much as possible and if you have a set left over, you check them.

Expanding on what @topsquark is saying, if [imath]m^2+n^2=40[/imath] then it's divisible by 4, so both [imath]m,n[/imath] must be even so trial and error [imath]\{\pm0,\pm 2,\pm 4,\pm 6\}.[/imath]

 

[AvgStudent]

Awesome, thanks!