If $a+b+c+d = 4m+1$ where $m$ is a positive number and $a , b,c,d \in \mathbb Z$ , then the minimum value of $a^2+b^2+c^2+d ^2$ is $4m^2 +2m +1$. If $a+b+c+d = 4m+1$ where $m$ is a positive number and $a , b,c,d \in \mathbb Z$ , then the minimum value of $a^2+b^2+c^2+d ^2$ is $4m^2 +2m +1$.
How to prove it without using calculus?
I know if $a , b,c,d \in \mathbb R$ then $a^2+b^2+c^2+d ^2$ will be minimum when $a =b= c =d = \frac{4m+1}{4}$..So the minimum value would have been $4m^2 +2m +1/4$..But what to do in that case?

The nearest integer greater than $4m^2 +2m +1/4$ is $4m^2 +2m +1$..And if we consider $a=b=c =m $ and $d=m+1$.We can get this minimum value.
