12
$\begingroup$

Find the number of triplets $(a,b,c)$ satisfying $(a^3+b)(b^3+a)=2^c$, where $a,b,c\in \mathbb{N}$

A trivial solution is $(1,1,2)$. I think there aren't any other such triplets, so I've been trying to prove that but haven't been able to do so.
How can other solutions be found?

$\endgroup$
  • $\begingroup$ The source of the problem? $\endgroup$ – hrkrshnn Oct 20 '14 at 9:04
  • $\begingroup$ @Moronplusplus I have no idea, but I'll try to find out. $\endgroup$ – Shaurya Gupta Oct 20 '14 at 16:27
9
+50
$\begingroup$

shaurya gupta already noted the solution $(a,b,c) = (1,1,2)$; rogerl found another pair, $(3,5,12)$ and $(5,3,12)$, and conjectured that there are no others. I prove that this conjecture is correct.

[The question required that $a,b,c \in \mathbb N$; this notation "$\mathbb N$" is sometimes used for nonnegative integers, but I assume that zero is not allowed here, else $(a,b,c) = (0,2^r,4r)$ is a solution for every nonnegative integer.]

rogerl shows that $a,b$ must be odd. Assume without loss of generality that $a \leq b$, and write $$ a^3 + b = 2^d, \quad b^3 + a = 2^e $$ for some integers $d,e$ with $d \leq e$ and $c = d+e$. Then $b = \lfloor 2^{e/3} \rfloor$, which makes it easy to check that the three known solutions are the only ones with $e \leq 15$ (for each $e$, solve for $b$, recover $a = 2^e - b^3$, and check whether $0 < a \leq b$). Thus any other solution must have $e>15$, and therefore $d>5$ (because $b^3+a < (a^3+b)^3$).

Now $a,b$ satisfy $a^3 \equiv -b$ and $b^3 \equiv -a$ mod $2^d$. Therefore $a^9 \equiv a \bmod 2^d$. Since $a$ is odd it follows that $a^8 \equiv 1$, so $a \equiv \pm 1 \bmod 2^{d-3}$. Hence $a=1$, else $a \geq 2^{d-3}-1$ and $a^3 > 2^d$. But then $b=2^d-1$, and (since $d>5$) we have reached a contradiction because $b^3+a$ is strictly between $2^{3d-1}$ and $2^{3d}$. QED

P.S. For the record:

@ $a,b$ are odd because if a power of $2$ is the sum of two positive integers then their 2-valuations are equal, and this would not be possible with $a,b$ even.

@ One proof of the implication $a^8 \equiv 1 \bmod 2^d \Longrightarrow a \equiv \pm1 \bmod 2^{d-3}$ is to write $\pm a = 1 + 4n$ and expand $(a^8-1)/32$ as a polynomial in $n$.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ I'm sure this is obvious, but why does $e>15$ imply $a>2^5$? $\endgroup$ – rogerl Oct 23 '14 at 18:29
  • $\begingroup$ Sorry. Good question; it doesn't (else I could have been done at $\;a=1$). It does, however, imply $\;d>5$. Let me fix this... $\endgroup$ – Noam D. Elkies Oct 23 '14 at 19:11
6
$\begingroup$

Note that $a$ and $b$ must be odd and relatively prime. For they must have the same parity, and if they are both even, say $a = 2^rt$ and $b = 2^su$ for $t$ and $u$ odd, then (assuming wlog that $1\le r\le s$) $$(a^3+b)(b^3+a) = (2^{3r}t + 2^su)(2^{3s}u+2^rt) = 2^r(2^{3r}t + 2^su)(2^{3s-r}u+t) = 2^c.$$ Thus $(2^{3r}t + 2^su)(2^{3s-r}u+t)$ is a power of $2$, so that $3s-r=0$ and $r=3s$. But then the first factor cannot be a power of $2$.

The only solution with $a, b\le 10,\!000$ is $\{3,5,12\}$ (and of course $\{5,3,12\}$).

Given the relative paucity of cubes, I'd be surprised to find additional solutions.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ How will other solutions be found? How did you get the $\le1000$ limit? $\endgroup$ – Shaurya Gupta Oct 19 '14 at 13:06
  • $\begingroup$ I got it by brute forcing the problem. $\endgroup$ – rogerl Oct 19 '14 at 13:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.