This will be a proof by contradiction. Let us assume that $b$ is not equal to 0. Since $c$ is even this means that $c^5$ is also even. However, since $4b^5$ is even, this means that $a^5$ must also be even. Since $a^5$ is even, this means that $a$ is even. Now, let $a=2a_0$ and $c=2c_0.$ The equation now turns into $32a_0^5+4b^5=32c_0^5.$ However, since all the terms are divisible by 32, this means that $4b^5$ is also divisible by 32, which implies that $b$ must also be divisible by 2. Now, let $b=2b_0.$ We can continue substituting like this to get an infinite equation. Since $b$ is not equal to zero, this means that $a=b=c=\infty.$ Thus, this means that our previous claim is wrong and we can conclude that $b$ must be equal to 0 for c to be even.

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