Find all possible values of a-b

[anemone]

The numbers $a$ and $b$ are prime and satisfy $\dfrac{a+1}{a}+\dfrac{b}{b+1}=\dfrac{2k}{k+2}$ for some positive integer $k$.

Find all possible values of $a-b$.

 

[anemone]

Hint:

Spoiler

Notice that $a$ and $b+1$ are coprime.

 

[kaliprasad]

Hint:

Spoiler

Notice that $a$ and $b+1$ are coprime.

Not solved since long

Spoiler

above assumption is incorrect as a= 3, b= 5. if above is true then we need to prove it

 

[anemone]

Solution of other:

Spoiler

$\dfrac{a+1}{a}+\dfrac{b}{b+1}=\dfrac{2k}{k+2}$

Subtract 2 from both sides to get

$\dfrac{1}{b+1}-\dfrac{1}{a}=\dfrac{4}{k+2}$

From this, since $k$ is positive, we have that $a>b+1$. Therefore $a$ and $b+1$ are coprime, since $a$ is prime.

Group the terms on the LHS to get

$\dfrac{a-b-1}{a(b+1)}=\dfrac{4}{k+2}$

Now, $(a,\,a-b-1)=(a, b+1)=1$ and $(b+1,\,a-b-1)=(b+1,\,a)=1$ so the fraction on the left is in lowest terms.

Therefore the numerator on the left must divide the numerator on the right, which is 4. Since $a-b-1$ is positive, it must be $1,\,2$ or $4$ so that $a-b$ must be $2,\,3$ or $5$. All of these can be attained by $(a,\,b,\,k)=(5,\,3,\,78),\,(5,\,2,\,28)$ and $(7,\,2,\,19)$ respectively.