<s>[INST] Finding all integers such that $a^2+4b^2 , 4a^2+b^2$ are both perfect squares Are there any integers $a,b$, such that:
$$a^2+4b^2 , 4a^2+b^2$$
are both perfect squares?
 [/INST] Integers $m$ such that,
$$ma^2+b^2=d^2\\a^2+mb^2 = c^2\tag1$$
have solutions unfortunately do not have an official name. However, the $n$ of a similar system,
$$a^2+b^2 = c^2\\a^2+nb^2=d^2\tag2$$
are called concordant numbers and it is well-established it has positive solutions $a,b$ for the sequence,
$$n=1, 7, 10, 11, 17, 20, 22, 23, 24, 27, 30, 31, 34, 41, 42, 45, 47, 49,\dots$$
Let $m=r^2,$ and $a=p/r,\; b= q$, then,
$$p^2+q^2=s_1^2\\p^2+r^4q^2 = s_2^2\tag3$$
But since in your case we have $m=r^2=4,$ and $n=r^4=16$ is not a concordant number, then $(1)$ has no positive integer solutions.
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