Triplets satisfying $(a^3+b)(b^3+a)=2^c$

Solution 1:

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 $n$.

Solution 2:

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.