<s>[INST] For which primes $p$ are there integers $a,b,q$ so that $pq^2=a^4+4b^4$? There is a right triangle with rational sides and are $p$ iff there exists a solution.
$$pq^2=a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)$$
Without loss of generality, choose $a$ odd and $a,b$ coprime, then $a^2+2b^2+2ab$ and $a^2+2b^2-2ab$ are also coprime and $p\equiv 1 \pmod 4$. Divide into two cases:


*

*$a^2+2b^2+2ab=pk_1^2$, $a^2+2b^2-2ab=k_2^2$

*$a^2+2b^2+2ab=k_1^2$, $a^2+2b^2-2ab=pk_2^2$


These are the same under the transformation $a\to-a$, so I can ignore the second. Changing $a\mapsto a+b$ the system becomes:
$$a^2+b^2=pk_1^2$$
$$(a-2b)^2+b^2=k_2^2$$ 
There is no solution for $p=17$, and none for $p=29,61,73,89$ with $a,b<10^5$.
 [/INST] Unless I am getting really stupid in my old age, a rational triangle with area $p$ means that $p$ will be a Congruent Number. The first prime examples are $p=5$ and $p=7$. But a solution of $pq^2=a^4+4b^4$ must have $p \equiv 1 (\bmod 4)$, as the OP states. Thus the "iff" statement in the first sentence must be wrong.
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