The definition of a pseudo-square in this case is let $n=pq$ where $p$ and $q$ are primes. A pseudo-square mod $n$ will be defined as a number $a$ such that $(\frac{a}{p}) = (\frac{a}{q}) = -1$ (Legendre symbol) so that $(\frac{a}{n}) = 1$ (Jacobi Symbol) although a is not congruent to a square mod $n$
So, the first question asks me to find all psuedosquares mod 15.
So i can start out by writing 15 into its prime factorization, 5*3. So now i have
$(\frac{a}{3}) = (\frac{a}{5}) = -1$
It seems that the obvious way to do this would test everything mod 3 so test $(\frac{\{1,2\}}{3})$ The only number that will give me $-1$ in this case will be $(\frac{2}{3})$. Thus the only possible $a$ will be 2. So now I test $(\frac{2}{5})$ and see that gives me $-1$.
Now I test $(\frac{2}{15})$, well, $15\equiv 7$ mod 8 so $(\frac{2}{15}) = 1$
Thus $2$ is the only psuedosquare mod $15$
My main question is, imagine if I am given a larger number n, for example 187, which is factored 11*17. Is there a way to speed up this process without having to manually check every combination?
Also, how can I know for sure that $(\frac{2}{15}) = 1$ means 2 is not a square congruent to 15. I know that by definition of the jacobi symbol, $1$ does not imply that 2 is a square mod 15. But -1 does imply that its not. Do i manually have to check that 2 is not a square congruent to 15 somehow, or does the prereq that $(\frac{a}{3}) = (\frac{a}{5}) = -1$ automatically imply that its not.
As pointed out by Daniel Fischer in his comment, determining which $a$ satisfies $(\frac{a}{p})=(\frac{a}{q})=-1$, only hard work will do the trick.
As to your other question, the interesting answer is that you don't have to check whether the resulting set of $a \mod pq$ are quadratic residues or not: they never are!
This follows easily from the fact that $(\frac{a}{p})=(\frac{a}{q})=-1$: suppose that $a$ is a quadratic residue $\mod pq$, then it follows that it also is a quadratic residue $\mod p$ and $\mod q$, which means that $(\frac{a}{p})=(\frac{a}{q})=+1$, a contradiction.
Also note that $(\frac{a}{pq})=+1$ is a corollary of $(\frac{a}{p})=(\frac{a}{q})=-1$, as $(\frac{a}{pq})$ equals $(\frac{a}{p})(\frac{a}{q})$.