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EDIT: Please see EDIT(2) below, thanks very much.

I want to prove by infinite descent that the positive divisors of integers of the form $a^2+3b^2$ have the same form. For example, $1^2+3\cdot 4^2=49=7^2$, and indeed $7,49$ can both be written in the same form: $49$ is already shown, and $7$ is $2^2+3\cdot 1^2$.

I'm looking for a general proof of this using infinite descent. I tried to let $a^2+3b^2=xy$, where $x,y$ are positive integer factors of the LHS, but couldn't continue from here I'm afraid. However, I have managed to prove that integers of the form $a^2+3b^2$ are closed under multiplication, i.e. the product of two of them is of the same form. I was hoping this would help.

I also had another question if you have time: Is the representation in the form $a^2+3b^2$ unique? For example, is it possible for an integer, say $N$, to be $N=a^2+3b^2=x^2+3y^2$ with $a\ne x$ and $b\ne y$?

EDIT: Sorry, here is the "closed under multiplication" proof I found:

Using Diophantus' Identity we have $(a^2+3b^2)(x^2+3y^2)=(a^2+(\sqrt{3}b)^2)(x^2+(\sqrt{3}y)^2=(ax+\sqrt{3}\cdot \sqrt{3} y)^2)+(a\cdot \sqrt{3} y - \sqrt{3} b\cdot x)^2=(ax+3by)^2+3(ay-bx)^2$ as required.

EDIT (2): I am very sorry. The question should be: If $\gcd(a,b)=1$, then every odd factor of $a^2 + 3b^2$ has this same form.

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  • $\begingroup$ Note that if $p$ is not of this form then using $pa$ and $pb$ gives a number with a factor $p$, so you have to have some conditions on $a$ and $b$ to exclude such cases. $\endgroup$ – Mark Bennet Aug 5 '14 at 16:19
  • $\begingroup$ But $a=b=1$ is still a counterexample, $(a,b)=1$ and $2$ cannot be expressed as $x^2 + 3y^2$ $\endgroup$ – Darth Geek Aug 5 '14 at 21:33
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EDIT: added second answer proving the main question.

Since we've proven that the family of numbers in the form of $a^2+3b^2$ is closed unfer product, we just jave to prove that any prime divisor $d$ of $a^2 + 3b^2$ can be represented as $r^2 + 3s^2$ for some integers $r,s$.

Let $d\ \vert \ (a^2 + 3b^2)$ then, $dk = a^2 +3b^2$ for a certain $k$.

If $d=1$ or $k = 1$ then the representation is obvious.

($d = 1^2+3·0^2$ for the first case and $d = a^2 + 3b^2$ for the second)

By integer division:

$$a = md + r \hspace{2cm} b=nd+s$$

If $r >d/2$ then $a = (m+1)d + (r-d) = m'd + r'$ where $m' = m+1$ and $r' = r-d$ (yes, $r'$ is negative, that's fine, we just want that either $r < d/2$ or $-r' < d/2$

(we use strict inequalities since $d$ is odd and therefore $d/2 \notin \mathbb{Z}$)

Similarly, if $s > d/2$ we construct $n'$ and $s'$ acordingly.

(use $m', r'$ and/or $n', s'$ from here on if you constructed them)

So $$a^2 +3b^2 = (md+r)^2 + (nd+s)^2 = $$

$$m^2d^2+2mdr+r^2+3n^2d^2+6nds+3s^2 = d(m^2d+2mr+3n^2d+6ns)+r^2+3s^2$$

But $d\ \vert\ (a^2 + 3b^2)$, therefore $d\ \vert\ (r^2 + 3s^2)$

But $r < d/2$ and $s < d/2$ therefore $r^2 + 3s^2 < \displaystyle\frac{d^2}{4} + \frac{3d^2}{4} = d^2$, so $r^2 + 3s^2 = d$

(remember that $d$ is prime, so the only multiples of $d$ lower than $d^2$ are $d$ and $0$ and $r^2 + 3s^2\neq 0$ because otherwise $r = s = 0$ what would mean that $d$ divides both $a$ and $b$)

Thus any odd divisors, $d$ of $a^2 + 3b^2$ can be written as $r^2 + 3s^2$ for some integers $r$ and $s$.

NOTE: If you don't asume that $d$ is prime, then $d$ is a divisor of $r^2 + 3s^2 < a^2 + 3b^2$ and you can create the descent argument you wanted.

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  • $\begingroup$ Hi, this is an old answer but I wonder if I could seek some clarification - (remember that..) Why does d being prime mean that multiples of d other than d and 0 aren't allowed as a value for the expression in question? $\endgroup$ – AnotherJohnDoe Aug 6 '17 at 15:27
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    $\begingroup$ @AnotherJohnDoe I should proofread my proofs before posting them. This was from three years ago (to the day!). Just glancing over it I can see that there are some typos and I wouldn't be surprised to find some mistakes (how embarrasing!). Let me check it a bit further and I'll edit the answer with either a detailed explanation or an alternative reasoning. $\endgroup$ – Darth Geek Aug 6 '17 at 17:36
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Note that: $$(ax-3by)^2 + 3(ay+bx)^2 = (a^2+3b^2)(x^2+3y^2) = (ax+3by)^2 + 3(ay-bx)^2$$

So if all divisors of numbers of that form are also of that form, then only numbers of that form that are primes have unique representation.

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This is false as stated. $4$ has an expression as $a^2 + 3 b^2,$ with $a=b=1,$ but $2$ does not. $25$ has such an expression, with $a=5,b=0,$ but $5$ itself does not. The same applies to any $q^2,$ where $q \equiv 2 \pmod 3$ is prime.

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For your second question. Not unique. $$28=4^2+3\cdot 2^2=5^2+3\cdot 1^2=1^2+3\cdot 3^2.$$ $$84=6^2+3\cdot 4^2=3^2+3\cdot 5^2.$$

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  • $\begingroup$ @EulerianAdventurer: I found another example:) $\endgroup$ – mathlove Aug 5 '14 at 16:25
  • $\begingroup$ @EulerianAdventurer: haha, I hope so:) $\endgroup$ – mathlove Aug 5 '14 at 16:49

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