Xrfgtjtk

Here's the problem I'm having difficulties with:

Find all positive integers $a$ and $b$ such that $$(1 + a)(8 + b)(a + b) = 27ab,.$$

Does anyone have an idea how to do this? Any detailed solution is welcome! :)

Using Hölder's inequality,
$$27ab = (a+1)(8+b)(b+a) geqslant left(2sqrt[3]{ab}+sqrt[3]{ab} right)^3=27ab$$

Hence we are looking for the equality case for Hölder, which is when $a:8:b=1:b:a implies (a, b)=(2, 4)$.

In fact, this is the only solution among positive reals, not just positive integers.

This is a supplementary solution, where I solve for all $(a,b)inmathbb{Z}timesmathbb{Z}$ such that $$(1+a)(8+b)(a+b)=27ab,.$$ From $(1+a)(8+b)(a+b)-27ab=0$, we have
$$(8+b)a^2+big((8+b)(b+1)-27bbig)a+b(8+b)=0,.$$
The discriminant of this quadratic polynomial with respect to $a$ is
$$begin{align}big((8+b)(b+1)-27bbig)^2-4cdot(8+b)cdot b(8+b)&=b^4-40b^3+276b^2-544b+64\&=(b-4)^2(b^2-32b+4),.end{align}$$
We require that $(b-4)^2(b^2-32b+4)$ be a perfect square. If $b=4$, then $$12(a-2)^2=12left(a^2-4a+4right)=0,,$$ so $a=2$. If $bneq 4$, then $$(b-16)^2-252=b^2-32b+4=c^2$$
for some integer $c$. Thus, $$d^2-c^2=252,,$$ where $d:=b-16$.

Since $4mid 252$ but $8nmid 252$, both $c$ and $d$ are even. Let $c:=2p$ and $d:=2q$, so that $$(q+p)(q-p)=q^2-p^2=frac{d^2-c^2}{4}=63tag{*},.$$
Therefore, the possible values of $(q+p,q-p)$ are $$(-63,-1),,,,(-21,-3),,,,(-9,-7),,,,(-7,-9),,,,(-3,-21),,,,(-1,-63),,$$ $$(1,63),,,,(3,21),,,,(7,9),,,,(9,7),,,,(21,3),,text{ and }(63,1),.$$
Thus, $b-16=d=2q=(q+p)+(q-p)$ takes the $6$ values
$$-64,-24,-16,+16,+24,+64,.$$
Ergo, $bin{-48,-8,0,32,40,80}$, resulting in the following solutions $(a,b)$:
$$(80,-48),,,,(0,-8),,,,(-1,0),,,,(0,0),,,,(-5,32),(-16,40),,text{ and }(-55,80),,$$
as well as the pair $(2,4)$ found earlier.

Using (*), we can also find all rational solutions. By setting $r:=q+p$, the rational solutions $(a,b)neq (2,4)$ take the form
$$left(-frac{(3+r)(7+r)}{21+r},frac{(7+r)(9+r)}{r}right)text{ for }rinmathbb{Q}setminus{0,-21},,tag{#}$$
and
$$left(-frac{(9+r)(21+r)}{r(3+r)},frac{(7+r)(9+r)}{r}right)text{ for }rinmathbb{Q}setminus{0,-3},.tag{@}$$
By the way, I just realized that with the transformation $rmapstodfrac{63}{r}$, the two solutions (#) and (@) are identical. (The same parametrization also works if you want to solve for real solutions $(a,b)neq (2,4)$, or even complex solutions $(a,b)$, where $r:=-6pm3sqrt{3}text{i}$ gives rise to the pair $(a,b)=(2,4)$.)

This is not a complete solution, but it points the way to one, and its simplicity, I think, makes it worth mentioning.

By expanding out the product and rearranging the results, we obtain the equivalent equation to solve,

$${8+8aover b}+{8+bover a}=18-a-b$$

Since the left hand side is positive, the right hand side limits the possibilities for $a$ and $b$ to a small enough set for brute force to take over.