Let $w, x, y, z$ be natural numbers. Find the correct alternative. Let $w$, $x$, $y$, $z$ be four natural numbers such that their sum is $8\cdot m+10$, where $m$ is a natural number. Given $m$ which of the following is possible:


The max. possible value of $w^2+x^2+y^2+z^2$ is $6\cdot m^2+40\cdot m+26$.
The max. possible value of $w^2+x^2+y^2+z^2$ is $16\cdot m^2+40\cdot m+28$.
The min. possible value of $w^2+x^2+y^2+z^2$ is $16\cdot m^2+40\cdot m+28$.
The min. possible value of $w^2+x^2+y^2+z^2$ is $16\cdot m^2+40\cdot m+26$.


The answer given is option 4.

I tried to expand $(w+x+y+z)^2$. This will be $0\ (\operatorname{mod}\ 4)$. Now the only way sum of four integers give even number is if 1. All even, 2. All odd, 3. 2 odd and 2 even. For the first 2 cases if we expand the expression everything except $w^2+x^2+y^2+z^2$ will be a multiple of four. So, $w^2+x^2+y^2+z^2$ should be divisible by $4$. Only case 3 requires $w^2+x^2+y^2+z^2$ to be divisible by $2$. Now if we put $m=1$ and take $w, x, y, z$ as $1, 3, 5, 9$ we can rule out first 2 options of the question. From here, can we argue that option 4 is the possible choice? Since option 4 gives $2\ (\operatorname{mod}\ 4)$ case. 

Is there any other general way to prove this? This is a question from a chapter of number theory. But solution in any form is heartily welcome.
You can use the AM-QM inequality;

$$ 2m + \frac 52= \frac {w + x + y + z}4 \le \sqrt{\frac{w^2 + x^2 + y^2 + z^2}4}$$

Hence you easily get $w^2 + x^2 + y^2 + z^2 \ge 16m^2 + 40m + 25$

Your book reports $26$ instead of $25$; this can be fixed by noting that to have equality you should have $w = x = y = z$, but his would imply $8m + 10 = 4w$. The left hand side is not divisible by $4$ though, so we cannot have equality and the minimun attainable is at lest one above the previous bound, that is to say $16m^2 + 40m + 26$.

Of course one should now prove that you can actually reach this; you can do so by following the general line guide "the more the numbers are close to each other, the less their quadratic mean will be" (if the arithmetic mean is fixed, of course)

So following this idea you can try to set $x = w = 2m + 2$, $y = z = 2m + 3$. You can easily see that with this choice $$\begin{cases} w + x + y + z = 8m + 10 \\ w^2 + x^2 + y^2 + z^2 = 16m^2 + 40m + 26\end{cases}$$ and it is indeed the minimum for what we have said before
The simplest aproach would be using Arithmetic Mean - Geometric Mean Inequality.

Let, $A=w+x+y+z = 8m+10$ . Therefore, $A^2 = 64m^2 + 160m +100 \; $

Also, $$A^2=w^2+2 w x+2 w y+2 w z+x^2+2 x y+2 x z+y^2+2 y z+z^2 $$
$$A^2 \le 4w^2 + 4x^2 + 4y^2 + 4z^2 \; \; \;   *$$ 

Therefore, $$w^2+x^2 + y^2 + z^2 \ge \frac{1}{4} A^2 = 16m^2 + 40m + 25$$
The equality is possible, if and only if $w=x=y=z=2m+2.5$ . 

Therefore, 


  $$min(w^2+x^2 + y^2 + z^2) = 16m^2 + 40m + 26$$




$* \; \; \; \;  \; 2wx \le w^2 + x^2, \; 2wy \le w^2 + y^2, \; and \; etc.$