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Let $a,b \in \mathbb{N}^{*}$. Prove that $\sqrt{13a^2+b^2}$ and $\sqrt{a^2+13b^2}$ cannot be simultaneously rational.

If $(a,b)=(k,k\cdot6)$, then $\sqrt{13a^2+b^2}$ is rational, but I do not know if those are the only solutions.

wythagoras
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user42768
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  • I interpret this question as "Show that $13a^2+b^2$ and $a^2+13b^2$ are not both square", rather than as a request to find when each is rational and showing that the two sets don't coincide. – Patrick Stevens Feb 14 '16 at 09:13

2 Answers2

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If both are rational then $14(a^2+b^2)$ is the sum of two squares. But it is impossible for (non-zero) $n$ and $14n$ to both be the sum of two squares as the well-known characterization requires the power of $7$ dividing each to be even.

Erick Wong
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    So the proof would be: Consider $m^2+n^2 = 14(a^2+b^2)$. Because the quadratic residues modulo 7 are {0,1,4,2}, then $m \equiv n \equiv 0 (mod 7)$, but that means that the RHS should be $\equiv 0 (mod 49)$, which is impossible? – user42768 Feb 14 '16 at 09:49
  • @user42768 why impossible? (you haven't justified it) To continue, $a^2+b^2\equiv 0\pmod{7}$, and again as you said $a\equiv b\equiv 0\pmod{7}$. Therefore $(m,n,a,b)$ is an integer solution if and only if $(m/7,n/7,a/7,b/7)$ is an integer solution, so by Infinite Descent $m=n=a=b=0$. But this is not the proof this answer was going for. To understand this answer, you need to know the theorem written here (last paragraph of the introduction). – user236182 Feb 14 '16 at 10:01
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    So, using the theorem you referenced, we know that $a^2+b^2$ is divisible by an even power of seven and multiplying it by 14 changes that. But because $14(a^2+b^2)$ can be written as a sum of squares, then it should be divisible by an even power of seven, which is a contradiction. Is this what this answer wanted? – user42768 Feb 14 '16 at 10:18
  • @user42768 yes. – user236182 Feb 14 '16 at 10:25
  • @user236182 Thank you, again. I am sorry for not being able to accept both answers. – user42768 Feb 14 '16 at 10:28
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$\sqrt{x}$ (where $x\in\mathbb Z^+$) is rational if and only if $\sqrt{x}$ is a positive integer (see this question).

Therefore, your problem is equivalent to proving the system $13a^2+b^2=k^2$ with $a^2+13b^2=m^2$ has no positive integer solutions.

$$a^2+13\left(k^2-13a^2\right)=m^2\iff m^2+168a^2=13k^2$$

I'll prove the last equation has no positive integer solutions.

$m^2\equiv 5k^2\pmod{8}$. But we have $y^2\equiv \{0,1,4\}\pmod{8}$ for all $y\in\mathbb Z$.

Therefore $m,k$ are both even. Let $m=2m_1$, $k=2k_1$. Then $m_1^2+42a^2=13k_1^2$.

$m_1,k_1$ have the same parity. If $m_1,k_1$ are both odd, then $m_1^2\equiv k_1^2\equiv 1\pmod{8}$, so $42a^2\equiv 4\pmod{8}$, i.e. $a^2\equiv 2\pmod{4}$, contradiction ($2$ is not a quadratic residue mod $4$). Therefore $m_1,k_1$ are both even, so $a$ is also even.

Therefore $(m,a,k)$ is a positive integer solution if and only if $(m/2,a/2,k/2)$ is also a positive integer solution. By Infinite Descent we get $m=a=k=0$.

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