MHB Uncovering Overlooked Solutions: Solving a Triple Integer Equation

[anemone]

Find all triples $(a,\,b,\,c)$ of positive integers such that $\dfrac{a}{b}+\dfrac{b}{c+1}+\dfrac{c}{a}=\dfrac{5}{2}$.

 

[Opalg]

Find all triples $(a,\,b,\,c)$ of positive integers such that $\dfrac{a}{b}+\dfrac{b}{c+1}+\dfrac{c}{a}=\dfrac{5}{2}$.

[sp]By the GM-AM inequality, $\Bigl(\dfrac{abc}{b(c+1)a}\Bigr)^{1/3} \leqslant \dfrac13\Bigl(\dfrac ab + \dfrac b{c+1} + \dfrac ca\Bigr) = \dfrac56.$

Therefore $\dfrac{abc}{b(c+1)a} \leqslant \Bigl(\dfrac56\Bigr)^3$. Taking reciprocals, $1 + \dfrac1c \geqslant \Bigl(\dfrac65\Bigr)^3 = 1.728.$ The only way that can happen for a positive integer $c$ is if $c=1.$

Putting $c=1$, the original equation becomes $\dfrac ab + \dfrac b2 + \dfrac1a= \dfrac52.$ Multiplying out the fractions, that becomes $2a^2 - b(5-b)a + 2b = 0$, a quadratic in $a$ with solution $a = \frac14\bigl(b(5-b) \pm\sqrt{b^2(5-b)^2 - 16 b}\bigr).$ For this to give a positive value for $a$ we must clearly have $b<5$. But $b=3$ and $b=4$ make that square root negative, so the only possible values for $b$ are $b=1$ and $b=2.$ It is easy to check that those both give solutions, with $a=b$ in both cases.

Therefore the solutions are $(a,b,c) = (1,1,1)$ and $(a,b,c) = (2,2,1)$.[/sp]

 

[anemone]

Thanks for participating, Opalg!:)

Solution of other:

Spoiler

We can rewrite the given equation to get:

$\dfrac{a}{b}+\dfrac{b}{c+1}+\dfrac{c}{a}=\dfrac{5}{2}$

$\dfrac{a}{b}+\dfrac{b}{c+1}+\dfrac{c+1-1}{a}=\dfrac{5}{2}$

$\dfrac{a}{b}+\dfrac{b}{c+1}+\dfrac{c+1}{a}-\dfrac{1}{a}=\dfrac{5}{2}$

$\dfrac{a}{b}+\dfrac{b}{c+1}+\dfrac{c+1}{a}=\dfrac{5}{2}+\dfrac{1}{a}$

Applying AM-GM inequality to the terms on the left, we have:

$\begin{align*}\dfrac{a}{b}+\dfrac{b}{c+1}+\dfrac{c}{a}&\ge 3\sqrt[3]{\dfrac{a}{b}\cdot \dfrac{b}{c+1}\cdot \dfrac{c+1}{a}}\\&\ge 3 \end{align*}$

This implies $\dfrac{1}{a}\ge \dfrac{1}{2}$, which means $a\le 2$.

We thus have to consider two cases, $a=2$ and $a=1$.

When $a=2$, equality must be attained, i.e. $\dfrac{2}{b}=\dfrac{b}{c+1}=\dfrac{c+1}{2}$.

It follows that $b(c+1)=4$ so we have two choices, $b=2$, $c=1$ or $b=1$, $c=3$.

However, note that the second pair of solution does not satisfy $\dfrac{2}{b}=\dfrac{b}{c+1}$, thus we get only one solution in this case, namely $(a,\,b,\,c)=(2,\,2,\,1)$.

If $a=1$, the equality becomes $\dfrac{1}{b}+\dfrac{b}{c+1}+c=\dfrac{5}{2}$, which implies $c<2$.

For $c=1$, we get $\dfrac{1}{b}+\dfrac{b}{2}=\dfrac{3}{2}$, i.e. $b^2-3b+2=0$. This gives two solutions, $b=1$ and $b=2$ that yield $(a,\,b,\,c)=(1,\,1,\,1),\,(1,\,2,\,1)$.

For $c=2$, we get $\dfrac{1}{b}+\dfrac{b}{3}=\dfrac{1}{2}$, i.e. $2b^2-3b+6=0$. This does not have positive integer solutions.

We conclude that all solutions are given by that yield $(a,\,b,\,c)=(1,\,1,\,1),\,(1,\,2,\,1),\,(2,\,2,\,1)$.

Opalg has overlooked the pair $(a,\,b,\,c)=(1,\,2,\,1)$ but I know that is purely an honest careless mistake.(Smile)

 

[Opalg]

Spoiler

Opalg has overlooked the pair $(a,\,b,\,c)=(1,\,2,\,1)$

[sp]Notice that the solution that I overlooked was in the section where I casually claimed "It is easy to check ...". Therein lies a lesson. (Shake)[/sp]