Proving all sufficiently large integers can be written in the form $a^2+pq$. This is one of those numerous questions I ask myself, and to which I seem unable to answer:

Can every integer greater then $657$ be written in the form $a^2+pq$, with $a\in\mathbb Z$ and $p,q$ prime?

A quick brute force check trough numbers up to $100000$ told me that the only positive integers that can not be represented in the given form are $1, 2, 3, 12, 17, 28, 32, 72, 108, 117, 297$ and $657$.
The above conjecture seems quite likely to me. Since pretty much integers are of the form $pq$, it is quite probable that, given a large $n>0$, at least one of $n, n-1, n-2^2, n-3^2, \ldots$ is of the form $pq$.
I.e, considering the set $F=\{4,6,9,10,14,15,21,22,25,26,\ldots\}$ of all numbers of the form $pq$, a number $n$ not satisfying the property implies that none of the integers $n, n-1^2,\ldots, n-\lfloor\sqrt n\rfloor^2$ is an element of $F$.
It's not much, but that's all I've discovered so far.
(Besides, is there any hope solving this riddle without using complicated analytic number theory?)

This is very hard.
(Notation: A semiprime is a product of exactly two distinct primes.)
Just consider this problem for perfect squares.  We're looking at the equation $b^2=a^2+pq$, which easily translates to $$(b-a)(b+a)=pq.$$  For fixed $b$, this is soluble in $a$ (and $p$ and $q$), precisely if either: 1) $2b-1$ is a semiprime, and we take $a=b-1$, or 2) There are primes $p$ and $q$ such that $p+q=2b$.  If we assume that $2b-1$ is not a semiprime (and this is true for most $b$ -- looking at $b<x$, there are $O(x \frac{\log\log x}{\log x})$ values of $b$ for which $2b-1$ is a semiprime), we're therefore reduced to solving the Goldbach problem, and I don't know how to say anything useful about that (though I wish I did!).  Thus, proving that every large integer $n$ can be written as $a^2+pq$ is at least as hard as proving Goldbach (or, I guess, proving Goldbach for a large subset of the even integers, those that aren't one more than a semiprime, but that's probably just as hard).
