MHB Prime Pairs: Finding $b-a$ Values

[anemone]

The numbers $a$ and $b$ are prime and satisfy $\dfrac{a}{a+1}+\dfrac{b+1}{b}=\dfrac{2k}{k+2}$ for some positive integer $k$. Find all possible values of $b-a$.

 

[Opalg]

[sp] Write it as $\Bigl(1-\dfrac1{a+1}\Bigr) + \Bigl(1+\dfrac1b\Bigr) = 2 - \dfrac4{k+2}$, which implies $\dfrac1{a+1} - \dfrac1b = \dfrac4{k+2}$. The right side of that last equation is positive, hence so is the left side. Therefore $b>a+1$.

Now multiply out the fractions, to get $(b-a-1)(k+2) = 4b(a+1)$. After playing around with that equation for a while, I decided that the best way to re-write it is as $$\bigl(b - (a+1)\bigr)\bigl(k - (4a+2)\bigr) = 4(a+1)^2.\qquad(*)$$ There are then two cases to consider.

Case 1: $a=2$. Then (*) becomes $(b-3)(k-10) = 36$. The only primes $b$ for which $b-3$ is a factor of $36$ are $5$ and $7$. These lead to the solutions $(a,b,k) = (2,5,28)$ and $(a,b,k) = (2,7, 19)$, the values of $b-a$ being $3$ and $5$.

Case 2: $a$ is an odd prime. Then $b - (a+1)$ is also odd. By (*), it is a divisor of $4(a+1)^2$, but it has no divisors in common with $4$ (apart from $1$). It also has no divisors in common with $a+1$ (apart from $1$), because any such divisor would be a nontrivial factor of $b$, contradicting the fact that $b$ is prime. The only remaining possibility is that $b - (a+1) = 1$, and then $b-a = 2$. There are (infinitely?) many such solutions, because $(a,b)$ can be any prime pair $(p,p+2)$.

In conclusion, the possible values of $b-a$ are $2,3,5$.[/sp]

 

[anemone]

Well done Opalg and thank you for participating in this challenge!

Solution provided by other:

Spoiler

Subtract 2 from both sides of the given equality yields

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

Since $k$ is positive, we must have $b>a+1$. Therefore, $b$ and $a+1$ are coprime, since $b$ is prime. Simplify the LHS we get $\dfrac{b-(a+1)}{b(a+1)}=\dfrac{4}{k+2}$.

Note that the fraction on the left is in lowest terms, therefore the numerator must divide the numerator on the right, which is 4.

Since $b-(a+1)>0$, it must be 1, 2 or 4 so that $b-a$ must be 2, 3 or 5.