Is this olympiad-like question about remainders an open problem?

This is a known theorem which is proved in this paper:

a(mod p) ≤ b(mod p) for all Primes p Implies a = b

Authors: P. Erdos, P. P. Palfy and M. Szegedy

The American Mathematical Monthly, Vol. 94, No. 2 (Feb., 1987), pp. 169-170.

Enjoy it!!!

(This is a comment, but is easier to enter as an answer.)

As Konstantinos Gaitanas pointed out, the result is in

a(mod p) ≤ b(mod p) for all Primes p Implies a = b

Authors: P. Erdos, P. P. Palfy and M. Szegedy

The American Mathematical Monthly, Vol. 94, No. 2 (Feb., 1987), pp. 169-170.

All of Erdos' publications are listed in https://www.renyi.hu/~p_erdos/Erdos.html and any of the papers can be downloaded from links there.

This particular paper is at https://www.renyi.hu/~p_erdos/1987-01.pdf

Some partial results:

By taking a large $p$, we see that $x\le y$. Let $d=y-x$. Assume $y>x$ (so certainly $y\ge 2$). If $p\mid y$ we obtain $x\bmod p\le y\bmod p=0$, hence $p\mid x$ and also $p\mid d$. Especially, $d\ge2$.

Now Let $p$ be a prime dividing $x+1$ (which is $\ge 2$). Then $x\bmod p=p-1$ implies $y\bmod p=p-1$, i.e. $p\mid y+1$. Hence $d$ is divisible by any prime dividing $x+1$. Especially, $d\ge 6$ because $x+1$ and $y$ have no prime in common.

Now let $p$ be any prime dividing $d+1$ (which is $\ge 7$). Then $y+1\bmod p=x\bmod p$ so that we can avoid the contradiction $y\bmod p<x\bmod p$ only by evading to $x\bmod p=0$, $y\bmod p=p-1$.

Now let $p$ be a prime dividing a number between $x$ and $y$, say $(r-1)p\le x<rp\le y$. Then $y-rp\ge y\bmod p\ge x\bmod p=x-(r-1)p$ implies $y\ge x+p$. Thus the $d$ consecutive numbers $x+1,\ldots, y$ are $d$-smooth (have only prime divisors $\le d$). In other words, ${y\choose d}\mid d!^m$ for suitable $m$