Example for $\dfrac{p_1p_2-1}{p_1+p_2}$ being odd natural number . If $p_1,p_2$ are odd prime numbers , is it possible that  
$\dfrac{p_1p_2-1}{p_1+p_2}$ is odd natural number greater than 1.
 A: Hint: Suppose $\dfrac{p_1p_2-1}{p_1+p_2} = 2k+1$ for some positive integer $k$. Then, we get:
$p_1p_2-1 = (2k+1)(p_1+p_2)$
$p_1p_2-(2k+1)p_1-(2k+1)p_2-1 = 0$
$p_1p_2-(2k+1)p_1-(2k+1)p_2+(2k+1)^2 = (2k+1)^2+1$
$(p_1-(2k+1))(p_2-(2k+1)) = 2(2k^2+2k+1)$
Clearly, $2k^2+2k+1$ is odd, so exactly one of $p_1-(2k+1)$ and $p_2-(2k+1)$ can be even. 
Do you see how this helps?
A: Exactly one of $p1+p2,p1p2-1$ is a multiple of 4.
A: Write the odd numbers $p_1,p_2$ and the result als numbers of the form $2k+1,2j+1,2m+1$
Then
$$ {(2k+1)(2j+1)-1 \over 2k+1+2j+1} = 2m+1 \tag 1$$
Expanded
 $$  {4kj + 2(k+j) \over 2k+2j+2} = 2m+1 \\
  2kj + (k+j) = (2m+1)(k+j+1) \\
  2kj + (k+j) = (2m+1)(k+j)+(2m+1) \\
  2kj  = 2m(k+j)+2m+1 \\
 2kj  = 2m(k+j+1)+1 $$                      
and finally
$$  2(kj-m(k+j+1))  = 1 \tag 2
$$
In the last row, the lhs is even and the rhs is odd, so no solution possible.      
Note,that the property of $p_1,p_2$ being prime was not used, only that they are odd numbers
A: $$\dfrac{p_1p_2-1}{p_1+p_2}=p_1-\left(\dfrac{p_1^2+1}{p_1+p_2}\right)$$ Therefore it is enough to find primes $p_1, p_2$ such that $$p_1^2+1|p_1+p_2.$$

One such way is, primes of the form $$p_2=p_1^2-p_1+1.$$

$(3,7),(7,43),(13,157)$ are the smallest such pairs.
Also I have no idea about the number of such prime pairs.
