# 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?

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What is an imperfect square? – celtschk Aug 20 '12 at 15:22
As usual: what have you tried? Is this homework? What is the motivation? – Old John Aug 20 '12 at 15:24
@celtschk "Perfect" squares emphasizes that they are squares of integers, as every positive real number is the square of some real number. – Ragib Zaman Aug 20 '12 at 15:24
@Old John i have possible solution , store2.up-00.com/Aug12/HNY76345.jpg – Frank Aug 20 '12 at 15:27
@MohammedAl-mubark: You should mention the trivial solutions. – Charles Aug 20 '12 at 15:28

Here are the first several steps. Notice that the equations are homogenous, so we can look for rational solutions and clear out the denominators at the end. The basic point here is that $\{ a^2+4b^2=c^2, 4a^2+b^2 = d^2 \}$ is a genus one curve in $\mathbb{P}^3$, and there are good methods to find rational points on genus one curves.
By the standard parametrization of Pythagorean triples, $(a,b,c)$ must be of the form $(x^2-y^2, xy, x^2+y^2)$. So we now want to solve $4 (x^2-y^2)^2 + (xy)^2 = d^2$. As in this question, set $u=x/y$ and $e=d/y^2$, so $4 (u^2-1)^2 + u^2 = e^2$. (Note that we have discarded the solutions at $y=0$.) We can recover $(x,y,d)$ from $(u,e)$ as $(x,y,d) = (uy, y, e y^2)$.
Expanding, we have $e^2=4u^4 - 7 u^2+4$ or $e^2 - (2u^2 -7/4)^2=15/16$. This is a genus one curve, with two rational points at $\infty$, where $e \approx \pm 2 u^2$. A genus one curve with a rational point is an elliptic curve. As in my previous answer, we now compute what that elliptic curve is explicitly. Set $$v = e-2u^2+7/4 \quad f = u(e-2u^2+7/4).$$ Note that $u=f/v$ and $e = v + 2 (f/v)^2 -7/4$, so $(u,e)$ are rational if and only if $(v,f)$ are. We have $$v(v+4(f/v)^2 - 7/2) = 15/16$$ or, clearing denominators, $$64 f^2 = 15 v + 56 v^2 - 16 v^3.$$