Let $Q(n,k)$ be the number of partitions of $n$ into $k$ distinct parts.
I want to show that for any $m \geq 1$ there exists $t \geq 1$ such that $$Q(n+t,i)=Q(n,i) \mod m \quad \forall n>0 \forall i\leq m$$
I have tried many many approaches, but I cant seem to prove it. I am pretty certain that it is true, since I have tested it computationally for many values.
My approaches have mainly been using the following identities/properties/facts:
A recurrence relation: $$Q(n,k) = Q(n-k,k) + Q(n-k,k)$$
From this I have also shown that
$$Q(n+t k,k) = \bigg(\sum_{i=0}^{t-1}Q(n+i k,k-1)\bigg) + Q(n,k)$$
Generating functions: I have tried using this fact: $$1 + \sum_{n,m \geq 1} Q(n,m) t^n u^m = \prod_{i \geq 1}(1+ut^i)$$
Induction proofs I have tried a bunch of induction proofs but nothing has worked.
I have showed by hand that $$Q(n+4,i)=Q(n,i) \mod 2 \quad \forall n>0 \forall i\leq 2$$ and $$Q(n+18,i)=Q(n,i) \mod 3 \quad \forall n>0 \forall i\leq 3$$
But have had no luck in showing it for arbitrary $m$.
If anyone has a solution, a hint, an insight or anything, it would be greatly appreciated!