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

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! :)

• What kind of numbers are $a$ and $b$? Rationals? Integers? Nonnegative integers? Positive integers? – Batominovski Dec 8 '18 at 18:03
• Forgot to mention. I've edited the question. Thanks – Wolf M. Dec 8 '18 at 18:09
• What are your thoughts? Put your work there and upload it. – jayant98 Dec 8 '18 at 18:17
• From a quick glance, we have $(a+1)(a+b)>ab$, so $$b+8=\frac{27ab}{(a+1)(a+b)}<27\,.$$ Thus, $b\in\{1,2,\ldots,18\}$. This shouldn't be too hard now. If you run out of ideas, you can still check all the $18$ cases ($b=1,2,\ldots,18$), which is probably not too much work. There will be $18$ quadratic equations in $a$. – Batominovski Dec 8 '18 at 18:28
• From $LHS$ one knows $b$ divides $1+a$ or $8a$; and $a$ divides $8+b$ or $b$. – AdditIdent Dec 8 '18 at 18:32

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

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 $$(a,b)\in\mathbb{Z}\times\mathbb{Z}$$ such that $$(1+a)(8+b)(a+b)=27ab\,.$$ From $$(1+a)(8+b)(a+b)-27ab=0$$, we have $$(8+b)a^2+\big((8+b)(b+1)-27b\big)a+b(8+b)=0\,.$$ The discriminant of this quadratic polynomial with respect to $$a$$ is \begin{align}\big((8+b)(b+1)-27b\big)^2-4\cdot(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 $$12(a-2)^2=12\left(a^2-4a+4\right)=0\,,$$ so $$a=2$$. If $$b\neq 4$$, 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 $$4\mid 252$$ but $$8\nmid 252$$, both $$c$$ and $$d$$ are even. Let $$c:=2p$$ and $$d:=2q$$, so that $$(q+p)(q-p)=q^2-p^2=\frac{d^2-c^2}{4}=63\tag{*}\,.$$ 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, $$b\in\{-48,-8,0,32,40,80\}$$, 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 $$\left(-\frac{(3+r)(7+r)}{21+r},\frac{(7+r)(9+r)}{r}\right)\text{ for }r\in\mathbb{Q}\setminus\{0,-21\}\,,\tag{#}$$ and $$\left(-\frac{(9+r)(21+r)}{r(3+r)},\frac{(7+r)(9+r)}{r}\right)\text{ for }r\in\mathbb{Q}\setminus\{0,-3\}\,.\tag{@}$$ By the way, I just realized that with the transformation $$r\mapsto\dfrac{63}{r}$$, 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 $$r:=-6\pm3\sqrt{3}\text{i}$$ 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,

$${8+8a\over b}+{8+b\over a}=18-a-b$$

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.