Show that any non-negative rational integer may be written as the sum of two squares if it is of the form.... 
Show that any non-negative rational integer may be written as the sum of two squares if it is of the form $$2^ip_1^{k_1}p_2^{k_2}\cdots p_m^{k_m}q_1^{2j_1}q_2^{2j_2}\cdots q_n^{2j_n}$$ where the $p_1,p_2,\dots,p_m$ are all primes congruent to $1\bmod 4$, the $q_1,q_2,\dots,q_n$ are all primes congruent to $3\bmod 4$ and $i,k_1,k_2,\dots,k_m,j_1,j_2,\dots,j_n$ are all non-negative integers.

I am completely blank on how to form a solution, so help would be much appreciated.

first of all a rather trivial point: a square is a sum of two squares (of which usually one will need to be $0^2$)
now you need a lemma: if $a$ is a sum of two squares and $b$ is a sum of two squares, then the product $ab$ is also a sum of two squares. 
this is easy to see in terms of the multiplication of two Gaussian integers:
$$
(a^2+b^2)(c^2+d^2)=|a+ib|^2|c+id|^2=|(a+ib)(c+id)|^2=|(ac-bd)+i(ad+bc)|^2=(ac-bd)^2+(ad+bc)^2
$$
since $2=1^2+1^2$ the lemma implies any power of two is a sum of two squares. 
a well-known theorem tells us that any prime of the form $4n+1$ is the sum of two squares, so the lemma implies that any product of such primes $p_1^{k_1}p_2^{k_2}\cdots p_m^{k_m}$ is a sum of two squares.
finally to multiply a sum of two squares by a square (of the form $q_1^{2j_1}q_2^{2j_2}\cdots q_n^{2j_n}$) is again a sum of two squares.
the proof of the theorem about primes of the form $4n+1$ is a fairly easy consequence of two basic results: Wilson's theorem and the fact that the Gaussian integers are a Euclidean ring and therefore are a  Unique Factorization Domain. Wilson's theorem is a simple exercise in modular arithmetic. To demonstrate a division algorithm for the Gaussian integers is a little more technical but of an elementary nature. 
