TITLE: Is this olympiad-like question about remainders an open problem?
QUESTION [19 upvotes]: Suppose that we are given two positive integers $x$ and $y$ such that
$$x \mod p \leqslant y \mod p$$
for each prime number $p$. (Here, $x \mod p,\; y \mod p$ stand for the least non-negative residua.) Does it follow that $x = y$?
The problem is seemingly easy as we have to test $x,y$ against finitely many primes only. However, after several attempts I begin to wonder whether this is an open problem...
Note that it seems to be an open problem whether there is a prime number between a pair of squares (a reference would be appreciated), so the case where $x$ and $y$ are squares themselves is hard enough. However, it may well happen that this doesn't require such an argument.

REPLY [3 votes]: (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