Project Euler Problem 9: Special Pythagorean triplet
You may immediately

    return a * b * c;


as soon as the values are found.
The loop

    for (int cValue= 1; cValue&lt;=value/2; cValue++)


is superfluous: you already have a and b, and you know that \$a+b+c\$ must be equal to value, so

    cValue = value - aValue - bValue;


suffices.
Thou shalt not brute force.

Recall that the Pythagorean triple \$a, b, c: a^2 + b^2 = c^2\$ can be expressed as $$a = k(u^2 - v^2)\\
b = 2kuv\\
c = k(u^2 + v^2)$$
with coprime \$u,v\$. Since they must add up to a particular value \$N\$ (in your case, 1000), conclude that \$2k(u^2 + uv) = N\$, or \$v = \dfrac{N}{2ku} - u\$.


We know that sides must be positive, so we have 


\$u &gt; v \rightarrow u &gt; \dfrac{N}{2ku} - u \rightarrow u &gt; \sqrt{\dfrac{N}{4k}}\$, and 
\$v &gt; 0 \rightarrow \dfrac{N}{2ku} - u &gt; 0 \rightarrow u &lt; \sqrt{\dfrac{N}{2k}}\$

Now loop over the range [sqrt(125), sqrt(500)] looking or the divisors of 500 (the range is as small as [11 .. 22], and the only divisor is 20). I intentionally will not explain how to recover \$a, b, c\$ once the divisor is found. It is pretty straightforward.