For positive integers $a,b,c,d$ satisfying $(a+bc)(b+ca)=13^d$. Prove that d is even

For positive integers $a,b,c,d$ satisfying $(a+bc)(b+ca)=13^d$. Prove that d is even. $$$$ I've tried assuming d odd and looking for contradictions but still can't find it $$$$ I also tried expanding it but nothing new $$(a+bc)(b+ca)=ab+abc+b^2c+abc^2=13^d$$

Solution 1:

The equation being dealt with is

$$(a + bc)(b + ca) = 13^d \tag{1}\label{eq1A}$$

Since $a$, $b$, $c$ and $d$ are positive integers, then both $a + bc$ and $b + ca$ must be positive powers of $13$. Thus, have $a = 13^{e}(a')$ and $b = 13^{f}(b')$, where $a'$ and $b'$ have no factors of $13$. If $e \lt f$, then $a + bc = 13^{e}(a' + 13^{f-e}(b'))$, which is not a power of $13$. Similarly, $e \gt f$ also leads to a contradiction. Therefore, we have $e = f = m$, which gives

$$a = 13^m(a'), \; 13 \not\mid a', \; \; b = 13^m(b'), \; 13 \not\mid b' \tag{2}\label{eq2A}$$

Using this and dividing both sides of \eqref{eq1A} by $13^{2m}$ gives

$$(a' + b'c)(b' + a'c) = 13^{d'} \tag{3}\label{eq3A}$$

where $d' = d - 2m$ has the same parity as $d$. Next, since $a'$ and $b'$ are also positive integers, we get for some positive integers $s$ and $t$ that

$$a' + b'c = 13^s \tag{4}\label{eq4A}$$

$$b' + a'c = 13^t \tag{5}\label{eq5A}$$

with $s + t = d'$. Multiplying \eqref{eq4A} by $c$ and subtracting \eqref{eq5A} gives

$$\begin{equation}\begin{aligned} b'c^2 - b' & = 13^{s}c - 13^t \\ b'(c - 1)(c + 1) & = 13(13^{s-1}c - 13^{t-1}) \end{aligned}\end{equation}\tag{6}\label{eq6A}$$

Since $13 \not\mid b'$, then $c \equiv \pm 1 \pmod{13}$. If $c = 1$, then \eqref{eq4A} and \eqref{eq5A} results in $13^s = 13^t \implies s = t$, with this giving $d'$ and $d$ being even. Otherwise, using the $p$adic order function, if $\operatorname{ord}_{13}(c \mp 1) = k$, then $c = 13^{k}u \pm 1$ where $13 \not\mid u$. Substituting this into \eqref{eq4A} and \eqref{eq5A} gives

$$a' \pm b' + b'(13^{k}u) = 13^s \tag{7}\label{eq7A}$$

$$b' \pm a' + a'(13^{k}u) = 13^t \tag{8}\label{eq8A}$$

Next, note if $a' = b'$, then \eqref{eq7A} and \eqref{eq8A} results in $a' = b' = u = 1$ and $s = t = k$, so $d'$ and, thus $d$, are even. Otherwise, have

$$\operatorname{ord}_{13}(a' \pm b') = v \tag{9}\label{eq9A}$$

If $v \lt k$, then $\operatorname{ord}_{13}(a' \pm b' + b'(13^{k}u)) = \operatorname{ord}_{13}(b' \pm a' + a'(13^{k}u)) = v$, so $s = t = v$. Thus, $d' = 2v$ would be even, so $d$ would be even as well. If $v \gt k$, then $\operatorname{ord}_{13}(a' + b' + b'(13^{k}u)) = \operatorname{ord}_{13}(a' + b' + a'(13^{k}u)) = k$, so $s = t = k$ and, once again, we get that $d$ is even.

Thus, the only possibility where $d'$, and thus $d$, might possibly not be even is where $v = k$. Then \eqref{eq9A} gives, for some non-zero integer $w$ with $13 \not\mid w$, that

$$a' \pm b' = 13^k(w) \tag{10}\label{eq10A}$$

First, consider the case where it's $a' + b' \equiv 0 \pmod{13}$. Substituting \eqref{eq10A} into \eqref{eq7A} and \eqref{eq8A} gives

$$13^k(w + b'u) = 13^s \implies w + b'u = 13^{s-k} \tag{11}\label{eq11A}$$

$$13^k(w + a'u) = 13^t \implies w + a'u = 13^{t-k} \tag{12}\label{eq12A}$$

Note the powers of $13$ on the right side of \eqref{eq11A} and \eqref{eq12A} must both be positive. Thus, subtracting the $2$ equations gives

$$u(b' - a') \equiv 0 \pmod{13} \tag{13}\label{eq13A}$$

However, $13 \not\mid u$. Also, $b' + a' \equiv 0 \pmod{13}$ means $b' - a' \not\equiv 0 \pmod{13}$. Thus, \eqref{eq13A} can't be true.

With the second case of $a' - b' \equiv 0 \pmod{13}$, following similar steps will lead to $u(b' + a') \equiv 0 \pmod{13}$, with this giving another contradiction.

This means the assumption of $v = k$ can't be true. Thus, $s$ and $t$ must always be equal in \eqref{eq4A} and \eqref{eq5A}, leading to $d'$ and $d$ being even.

Note this proof only relies on $13$ being an odd prime, so any other odd prime in \eqref{eq1A} instead would also give the same result, i.e., that $d$ must be even.