# 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?

• What is an imperfect square? Aug 20, 2012 at 15:22
• As usual: what have you tried? Is this homework? What is the motivation? Aug 20, 2012 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. Aug 20, 2012 at 15:24
• @Old John i have possible solution , store2.up-00.com/Aug12/HNY76345.jpg Aug 20, 2012 at 15:27
• @MohammedAl-mubark: You should mention the trivial solutions. Aug 20, 2012 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, $$64 f^2 = 15 v + 56 v^2 - 16 v^3.$$

The next step is to find rational points on this elliptic curve. I know that there is standard software for this, but I don't know how to use it. Hopefully, someone how does will finish off this question (and the previous one).

• There are also tables where generators for the group of rational points are given for curves up to a given discriminant. Probably best to first put the curve in Weierstrass form, calculate its discriminant, and look it up somewhere. Aug 22, 2012 at 2:11

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.