EGMO 2012

$\frac{a}{a+1}+\frac{b+1}{b}=\frac{2n}{n+2} \ \ \Longrightarrow \ \ 1-\frac{1}{a+1}+1+\frac{1}{b}=2-\frac{4}{n+2}$ Or $\frac{1}{a+1}-\frac{1}{b}=\frac{4}{n+2} \ \ \Longrightarrow \ \ \frac{b-a-1}{b(a+1)}=\frac{4}{n+2}$ And finally $(n+2)(b-a-1)=4b(a+1)$ Notice that since $b > a+1$ and $a,$ $b$ are prime numbers, thus, $\gcd(b, a+1)=1$ . It follows that $b-(a+1)$ is co-prime to both $b$ and $a+1$ . Hence $b-a-1 \mid 4$ , $b-a-1=1, 2, 4$ and $b-a=2, 3, 5$ . Notice that $n=\frac{4}{b-a-1} b(a+1)-2$ is a positive integer.

First I divided the question in two parts where

Case I : $a=2$ which gives the solution as $(a,b,n)=(2,5,28),(2,7,19)$

Case II : $a,b$ are odd primes. $(a,b,n)=(p,p+2,2(2p^{2}+6p+3))$ where $p,p+2$ both are prime numbers.

For finding the solution of the case II, I wrote $b=a+m$ where $m$ is an even natural number. (since $a,b$ are odd primes and $b>a+1$ .)

Then I solved for $n$ in terms of $a,m$ .

$n=\frac{2(2a^{2}+2am+2a+m+1)}{m-1}$ $=\frac{4(a+1)^{2}}{m-1}+(4a+2)$

(Now see that $(4(a+1)^{2})=even,(m-1)=odd$ .

So unless $m=2$

$\frac{4(a+1)^{2}}{m-1}=non-integer$ and hence forth $n=non-integer$ ;which is false .

Therfore $m=2$ and we are done .

For case I after putting $a=2$ we get $b=\frac{3(n+2)}{n-10}=3+\frac{36}{n-10}$ .

Since $b$ is an integer we get $n=(11,12,14,16,13,19,22,28,46)$ .

But $b$ is prime .

Therefore $n=19,28$

We can rewriting equation to

$\frac{b-a-1}{b(a+1)}=\frac{4}{n+2}$

So it leads to two conditions

Case I:

$\begin{cases} k(b-a-1)=4 \text{-------(1)}\\ kb(a+1)=n+2 \text{-------(2)}\\ \end{cases}$

For some positive integer $k$ , because $RHS$ of $(2)$ must be positive.

So $k=1,2,4$ ,

When $k=1$ , $b-a=5$ , solving this we get a only solution which is $(2,7,19)$ proves $b-a=5$ is valid.

When $k=2$ , $b-a=3$ , solving this we can easily figure out a solution $(2,5,28)$ proves $b-a=3$ is valid.

When $k=4$ , $b-a=2$ , solving this we can easily figure out a solution $(3,5,78)$ proves $b-a=2$ is valid.

Case II:

$\begin{cases} (b-a-1)=4k \text{-------(3)}\\ b(a+1)=k(n+2) \text{-------(4)}\\ \end{cases}$

For some positive integer $k$ .

When $k=1$ , $b-a=5$ , don't need to prove this.

When $k=2$ , $b-a=9$ , we get $a=2 , b=11$ but $b(a+1)=33 \neq 2(n+1)$ so $b-a=9$ invalid.

When $k=4$ , $b-a=17$ , we get $a=2 , b=19$ but $b(a+1)=19 \times 3 \neq 4(n+1)$ so $b-a=17$ invalid.

So sum of the values of $(b-a)$ be $5+3+2 = \boxed {10}$

Feel free to tell me if I have wrong.