Ain't it Infinite?!

Firstly, we use the following identity, valid over all real numbers.

${(a+b+c+d+e)}^{2}=a^2+b^2+c^2+d^2+e^2+2(ab+ac+ad+ae+bc+bd+be+cd+ce+de)$

By putting values given in the question, we get the following two equations:

$a+b+c+d+e=40 , a^2+b^2+c^2+d^2+e^2=320$

YAY! What more do we need for the CS inequality. By the same, we have-

$(a^2+b^2+c^2+d^2+e^2)(1+1+1+1+1) \geq {(a+b+c+d+e)}^2$

And, indeed, it's the equality case! This means $a=b=c=d=e$ , and $a+b+c+d+e=40$ .

Thus, the only possible value of $a$ , and indeed, the sum of all possible values of $a$ , is $\boxed{8}$ .