Let's determine the integers $a$ and $b$ if $a^{4}+(a+b)^{4}+b^{4}$ is a perfect square.
I. solution. Suppose there exists an integer $x$ such that

$$
a^{4}+(a+b)^{4}+b^{4}=x^{2}
$$

Transform the equation: $a^{4}+b^{4}=x^{2}-(a+b)^{4}$. Further transforming the left side: $\left[(a+b)^{2}-2 a b\right]^{2}-2 a^{2} b^{2}=x^{2}-(a+b)^{4}$, which means $2(a+b)^{4}-4 a b(a+b)^{2}+2(a b)^{2}=x^{2}$. From this, $x^{2}=2\left[(a+b)^{2}-a b\right]^{2}$.

The equality can only hold if $x=0$, because on the left side, a square number stands, in which every prime factor appears with an even exponent, while on the right side, the exponent of 2 is odd.

Thus, $(a+b)^{2}-a b=0$. Completing the square, we get

$$
a^{2}+a b+b^{2}=0
$$

From this, if $b=0$, then $a$ is also equal to 0. If, however, $b \neq 0$, then

$$
a=\frac{-b \pm \sqrt{b^{2}-4 b^{2}}}{2}
$$

The discriminant is negative, so the equation has no solution even among the real numbers. The expression is only a square number when $a=b=0$.

II. solution. A square number (and thus any fourth power as well) when divided by 4 can only give a remainder of 0 or 1, the former when it is even, the latter when it is odd. Indeed, if $x$ is even, then $x=2 k$, so $x^{2}=4 k^{2}$ is divisible by 4. If, however, $x$ is odd, then $x=2 k+1$, so

$$
x^{2}=(2 k+1)^{2}=4 k^{2}+4 k+1=4\left(k^{2}+k\right)+1
$$

Thus, among $a, b$, and $a+b$, at most one can be odd, since their sum, which is a square number, can only give a remainder of 0 or 1 when divided by 4. However, if, for example, $a$ is odd, and then $b$ must be even, $a+b$ would also be odd, which is a contradiction. Therefore, $a$ and $b$ are both even (and then of course $a+b$ is also even): $a=2 a_{1}, b=2 b_{1}$. Therefore, if $a^{4}+(a+b)^{4}+b^{4}=16\left(a_{1}^{4}+\left(a_{1}+b_{1}\right)^{4}+b_{1}^{4}\right)$ is a square number, then $a_{1}^{4}+\left(a_{1}+b_{1}\right)^{4}+b_{1}^{4}$ is also a square number. Thus, $a_{1}$ and $b_{1}$ are both even, and so on. We have found that $a$ and $b$ are divisible by any power of 2, which is only possible if $a=b=0$; this is the only solution to the problem.