MHB Find all possible values of a-b

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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 $a-b$.
 
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Hint:

Notice that $a$ and $b+1$ are coprime.
 
anemone said:
Hint:

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

Not solved since long

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

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

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.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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