Solve $2a^2 + b^2 = 3c^2$ in $\mathbb{Z}$.

Question

I've been trying to solve this in the whole numbers

$$2a^2 + b^2 = 3c^2$$

I've tried basically all elementary methods, but couldn't get much. I do know that this equation should have infinitely many solutions, so eliminating anything by reducing modulo an integer is probably not gonna work. My current idea is to interpret the left hand side as the norm in $\mathbb{Z}[\sqrt{2}i]$, but I'm not sure where to go from that.

What about the standard method (for conics) of finding rational solutions to $2x^2+y^2=3$, via taking the line through $(1,1)$ with rational slope $m$ and finding the other intersection with the ellipse? — Daniel Schepler, Jun 21 '24 at 16:06
I understand this is Pell's Equation, and so maybe my suggestion isn't the most conventional. However, maybe it would be interesting approach would be to consider deriving infinitely many solutions via $x,y,z\in\mathbb{R}$ such that $x^2+y^2+z^2,$ and setting $a=\frac{x}{\sqrt{2}},b=y,c=\frac{z}{\sqrt{3}}.$ (Of course, you would need specifically find $x\in\mathbb{Z}[\sqrt{2}]-\mathbb{Z},y\in\mathbb{Z},z\in\mathbb{Z}[\sqrt{3}]-\mathbb{Z}.$ However, I feel that this shouldn't be too hard of a task when utilizing (informally called) SOH-CAH-TOA methods, though maybe I'm assuming too much.) — J.G.131, Jun 21 '24 at 16:37
@preferred_anon To be honest, no. That's the only solution that I've managed to find. — zaq, Jun 21 '24 at 17:16
"I've basically tried all elementary methods..." Are you familiar with the standard parametrization of all Pythagorean triples? The exact same method works here. — Servaes, Jun 21 '24 at 19:15
@Servaes The method only works partially here, though; the expressions derived that way, i.e., $(a,b,c) = (2q^2+2pq-p^2, p^2+4pq-2q^2, p^2+2q^2)$ don't provide all the primitive solutions--the primitive solutions must come from dividing through by their gcds. — Eric Snyder, Jun 22 '24 at 07:27
you can rewrite the equation this way: $a^2 + b^2 = 3c^2 - a^2$. and look for numbers that are sum of two squares that can also be written as $3c^2-a^2$. — user25406, Jun 22 '24 at 14:00
@zaq , see related post here : https://math.stackexchange.com/a/4885738/1296310 and https://math.stackexchange.com/a/4962185/1296310 — Guruprasad, Aug 28 '24 at 05:05

score 1 · Answer 1 · answered Jun 22 '24 at 19:32

Comment: some experimental results:

1- for any $n\in\mathbb z$ we can write:

$$3n^2=2n^2+n^2$$

2-A family of solutions results using this relation:

$$2n^2+(5n)^2=3(3n)^2; n\in\mathbb Z; a=n, b=5n, c=3n$$

This was resulted by writing:

$3\times 3^2=5^2+2\times 1^2$(primitive solution)

$3\times 6^2=10^2+2\times 2^2$

$3\times 9^2=15^2+2\times 3^2$

$3\times 12^2=20^2+2\times 4^2$

$3\times 25^2=5^2+2\times 5^2$

$\cdot$

Clearly more solutions come out by multiplying both sides by a perfect square.

3-We need to work on other groups such as:

$3\times 11^2=2\times 1^2+19^2$

$3\times 11^2=2\times 13^2+5^2$

$3\times 9^2=2\times 11^2+1^2$

$3\times 17^2=2\times 11^2+25^2$

$3\times 17^2=2\times 13^2+23^2$

$3\times 19^2=2\times 11^2+29^2$

$3\times 19^2=2\times 23^2+5^2$

$3\times 11^2=2\times 1^2+19^2$

All of these are primitive solutions and multiplying both sides by a perfect square make a new solution.

Eric Snyder · Accepted Answer · 2024-06-22T23:46:05.867

A partial answer, based on commentary above:

We can look for rational points on the ellipse $2x^2+y^2=3$. Essentially, we're setting $x = \frac a c, y = \frac b c$. The method is simple: a line with a rational slope passing through one rational point on a curve intersects at another rational point (assuming it intersects).

There is one obvious rational point: $(1,1)$. Because the ellipse is symmetrical, we can instead use $(-1,-1)$ as our initial rational point. Our line with rational slope $t$ is then $y= t(x+1)-1$. We can plug this in for $y$ in the ellipse's equation, and we get:

$$(2+t^2)x^2 + (2t^2-2t)x + (t^2 - 2t - 2) = 0$$

This can be solved by the quadratic formula, and the result is

$$(x,y) = \left(\frac{2+2t-t^2}{t^2+2},\frac{t^2+4t-2}{t^2+2}\right)$$

And since we set $c=1$, we can multiply through by the denominator to get

$$(a,b,c) = \left(2+2t-t^2,t^2+4t-2,t^2+2\right)$$

Now since $t$ is rational, we can substitute $t = \frac p q$ where $p,q \in \mathbb{N}$. Doing so, then multiplying through by $q^2$ (which appears in the denominators of all three), we get:

$$(a,b,c) = (2q^2+2pq-p^2, p^2+4pq-2q^2, p^2+2q^2)$$

This is a parametrized solution that gives integer solutions only. However, due to various quirks, it does not give only primitive solutions--if $p$ is even, or if $3 \mid p-q$, we end up with numbers that must be divided by their gcd to get a primitive solution. (We can see, for instance, that with $p$ even, $a,b,c$ will also all be even.) Additionally, if $\gcd(p,q) > 1$, it will find a non-primitive solution.

Here is a Desmos sheet that lets you play with $p,q$ and find various solutions. Note that if we get a negative solution, well, we're squaring it anyway, so you can just remove the negative sign.

Now, this ought to give all solutions. However, I'm uncertain that it does so, and don't have a proof. For instance, an answer linked above gives a different parametrization:

$$(a, b, c) = (p^2 - 3q^2, p^2 + 6pq + 3q^2, p^2 + 2pq + 3q^2)$$

which also gives solutions that must be divided by the gcd to find primitive solutions. I haven't checked whether the two parametrizations give the same solution set.

Edit: Barring something weird happening with large $p,q$, these two parametrizations would appear generate two disjoint solution sets (using Sage). There may well be other parametrizations that produce other solution sets.

Edit 2: Neither of these solution sets contains $(1, 19, 11)$, which is noted as a solution in another answer. So these two parametrizations don't hit all solutions.

Solve $2a^2 + b^2 = 3c^2$ in $\mathbb{Z}$.

2 Answers2