MHB How Do You Solve POTW #287 Involving Integer Equations?

[anemone]

Here is this week's POTW:

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Positive integers $a,\,b, \,c, \,d$, and $e$ satisfy the equations

$(a + 1)(3bc + 1) = d + 3e + 1\\
(b + 1)(3ca + 1) = 3d + e + 13\\
(c + 1)(3ab + 1) = 4(26 − d − e) − 1$

Evaluate $d^2 + e^2$.

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[anemone]

Congratulations to castor28 for his solution(Smile), which you can find below:

Spoiler

Let us write $P$, $Q$, $R$ for the left hand side of the equations. Adding the three equations gives:

$P + Q + R = 117 \qquad(*) $

The last equation shows that $R = (c+1)(3ab+1)$ is odd. This shows that both factors are odd, and $c$ and at least one of $\{a,b\}$ are even.
Because $c$ is even, the first two equations give the congruences modulo 2:

$$
\begin{align*}
a+1 &\equiv d + e + 1 \pmod{2}\\
b + 1&\equiv d + e + 1 \pmod{2}
\end{align*}
$$
​

from which we conclude that $a\equiv b\pmod{2}$. Because of the previous remark, $a$, $b$, and $c$ are all even; in particular they are greater than or equal to $2$.

If we take $a=b=c=2$, we get $P=Q=R=39$ and $P+Q+R = 117$, which is the correct value. Since this is the smallest possible solution, and, as all the coefficients are positive, any other solution would make $P+Q+R$ greater, this is the only solution of $(*)$.

With these values, we get a linear system:

$$
\begin{align*}
\qquad d + 3e + 1&=39\\
3d + e + 13&=39
\end{align*}
$$

giving $d=5$, $e=11$, and $d^2+e^2=146$.

Alternative solution:

Spoiler

By expanding the three given equations and adding them all up we have
$9abc+3ab+3bc+3ca+a+b+c+3=117$

Multiply both sides of the equation above by 3 and rearranging the constant and finally factor the resultant equation gives
$27abc+9ab+9bc+9ca+3a+3b+3c+1=343$

This is equivalent to
$(3a+1)(3b+1)(3c+1)=7^3$

Since $a,\,b$ and $c$ are positive integers, this implies $a=b=c=2$ and therefore $d=5$ and $e=11$.

Hence, $d^2 + e^2=5^2+11^2=146$