A number can be factored into two factors whose difference is 6, and the sum of their fourth powers is 272. Which number is this?
I. Solution: If we denote the smaller factor by $x$, then the other factor is $x+6$, and according to the requirement,

$$
x^{4}+(x+6)^{4}=272
$$

However, our equation becomes more symmetrical if we choose the arithmetic mean of the factors as the unknown. Denoting this by $z$, the two factors are $z-3$ and $z+3$, so

$$
(z-3)^{4}+(z+3)^{4}=272
$$

and after rearranging, the odd powers of $z$ cancel out:

$$
z^{4}+54 z^{2}-55=0
$$

thus we arrive at a quadratic equation in $z^{2}$. The roots of this equation are $z_{1,2}^{2}=1$, giving $z_{1}=1$ and $z_{2}=-1$, while the other roots, $z_{3,4}^{2}=-55$, do not yield real values for $z_{3}$ and $z_{4}$. The factor pair corresponding to the root $z_{1}$ is -2 and +4, and the pair corresponding to $z_{2}$ is -4 and +2. The product in both cases is -8, which is the only (real) number that can meet the requirements of our problem. Indeed, $(-2)^{4}+4^{4}=(-4)^{4}+2^{4}=272$.

II. Solution: Let the number we are looking for be $n$, and the factors according to the problem be $x$ and $y$. Then, assuming $x>y$,

$$
\begin{gathered}
x-y=6 \\
x y=n \\
x^{4}-y^{4}=272
\end{gathered}
$$

From here, we can directly obtain an equation for $n$ if we write equations (3) and (4) as $x(-y)=-n, x^{4}+(-y)^{4}=272$. We notice that the left-hand sides of our equations are symmetric functions of $x$ and $-y$. Thus, the left-hand side of equation (2), which is also a symmetric function of $x$ and $-y$, can be transformed in such a way that only the left-hand sides of (2), (3), and (4) appear in it. Indeed, after expansion,

$$
x^{4}-4 x^{3} y+6 x^{2} y^{2}-4 x y^{3}+y^{4}-\left(x^{4}+y^{4}\right)-4(x-y)^{2} x y-2(x y)^{2}=1296
$$

so

$$
272-4 \cdot 6^{2} n-2 n^{2}=1296, \quad n^{2}+72 n+512=0
$$

and from this, $n_{1}=-8, n_{2}=-64$.

However, since in transitioning from (2) to (5) we did not perform an equivalent transformation, we need to check whether the obtained numbers satisfy the conditions of the problem. For this, we also need to determine the factors $x, y$. These can be calculated from (2) and (3). With $n_{1}$, we arrive at the same two factor pairs as in the first solution, while for $n_{2}$, there is no (real) factor pair, so we cannot consider it a solution.

Remark. Most contestants unjustifiably assumed that the factors are integers, and then, by trial and error in (1) or (4), they arrived at the same result. Some assumed only that the factors are rational and showed that (in this case) the factors are integers, and that the problem is over-determined, as the equation (2) is redundant because 272 can be written as the sum of the fourth powers of two integers in only one way. - However, according to the problem, neither the two factors nor the product itself need to be rational! Consider two counterexamples. Keeping (2) and (3) unchanged, let the sum of the fourth powers of the factors be different from the given problem, say: $x^{4}+y^{4}=386$; then, by similar calculations as before, $x=\sqrt{2}+3, y=\sqrt{2}-3, n=-7$ is obtained, so the number we are looking for is rational, but the factors are irrational. If, keeping (2) and (3) unchanged, $x^{4}+y^{4}=200$, then $x=\sqrt{\sqrt{748}-27}+3, y=\sqrt{\sqrt{748}-27}-3, n=\sqrt{748}-36$, so the number we are looking for, along with the factors, is irrational.