$x_1,x_2,\ldots,x_m$ is a permutation of $1,2,\ldots,m$ and $n_1,n_2,\ldots,n_m$ be integer and $m>1$ and $m$ is odd. Now, $f(x)$ be defined as $f(x)=\sum\limits_{i,j=1}^{m} n_ix_j$. If $a,b$ are distinct permutations, then prove that there exists $a,b$ such that $f(a)\equiv f(b)\mod m!$
My attempt:
I thought of using PHP and to show that two equal remainders exist. There are $m!$ available $f(x)$ as there are $m!$ ways of re-arranging $1,2,\ldots,m$ and there are $m!$ remainders available namely $0,1,\ldots,m!-1$.
Now, I cannot use PHP here as there are equal number of holes as that of pigeons. But, is there some remainders that do not occur, because it has to be true otherwise we will get a contradiction, but I cannot think of it and neither prove it. Please help.