Problem 955

1957 factored into primes is 103 $\cdot$ 19. Since 103 and 19 are relatively prime, the divisibility of the given expression by 1957 is proven if we separately prove the divisibility by the prime factors.

Our expression can also be written as:

$$\left(1721^{2 n}-73^{2 n}\right)-\left(521^{2 n}-212^{2 n}\right)$$

Both bracketed expressions are divisible by the difference of the bases. In the first bracketed part, the difference of the bases is $1648=16 \cdot 103$, and in the second, $309=3 \cdot 103$.

It can be seen that the entire expression is divisible by 103.

By grouping the expression differently:

$$\left(1721^{2 n}-521^{2 n}\right)+\left(212^{2 n}-73^{2 n}\right)$$

Due to the evenness of the exponents, both parts within the brackets are divisible by the sum of the bases, the first by $2242=118 \cdot 19$, and the second by $285=15 \cdot 19$.

Since the expression is also divisible by 19, we have thus proven its divisibility by 1957.

$$\text { Jenő Székely (Pécs, Nagy Lajos g. I. o. t.) }$$

Note: About 30 solvers tried to prove the theorem by testing its correctness for $n=1$, 2, and from this concluded that it is true for all $n$. We know that this conclusion does not hold. - Such "solutions" were, of course, not accepted.