Number Theory Question

[AvgStudent]

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

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

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

 

[topsquark]

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

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

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

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 $m^2+n^2=40$ then it's divisible by 4, so both $m,n$ must be even so trial and error $\{\pm0,\pm 2,\pm 4,\pm 6\}.$

 

[AvgStudent]

Awesome, thanks!