Diophantine equation

The equation can be rewrite as

$(2x+1)(2y-5) = 105 = 3 \times 5 \times 7$

Let $a, b$ be $(2x+1), (2y-5)$ respectively.

The number of divisors of $105$ is $(1+1)(1+1)(1+1) = 8$ . So, the number of pairs of $(a, b)$ is 16 (consider negatives), $x = 16$ .

The tricky part is the sum. For every $2x+1=k$ , there must exist another $2y-5=-k$ . For example, $(1,105)$ and $(-105, -1)$ . So we have

$\begin{aligned}2x+1+2y-5&= k-k\\ x+y &= 2\end{aligned}$

With this in hand, the sum of all pairs is $16*2$ .

Therefore, our desired answer is $32+16=48$

Rewrite the equation as

$(2x + 1)(y - \frac{5}{2}) + \frac{5}{2} = 55 \Longrightarrow (2x + 1)(2y - 5) = 105.$

We can factor $105$ as a product of $2$ integers in $16$ "ordered" ways, namely

$1*105 = 3*35 = 5*21 = 7*15 = 15*7 = 21*5 = 35*3 = 105*1 =$

$(-1)(-105) = (-3)(-35) = (-5)(-21) = (-7)(-15) = (-15)(-7) =$

$(-21)(-5) = (-35)(-3) = (-105)(-1).$

We then in turn equate $(2x + 1)$ and $(2y - 5)$ to the respective first and second elements of these $16$ products to obtain the following solution pairs $(x,y)$ :

$(0,55),(1,20),(2,13),(3,10),(7,6),(10,5),(17,4),(52,3),(-1,-50),$

$(-2,-15),(-3,-8),(-4,-5),(-8,-1),(-11,0),(-18,1),(-53,2).$

Adding up all $16$ $x_{i}$ and $y_{i}$ values and then adding $x = 16$ gives us a solution of $\boxed{48}.$

The problem itself lists the solutions as $(x_1,y_1),(x_2,y_2),.(x_n,y_n)$ and the upper limit of the summation is also $n$ . Therefore, logically the number of solutions should be designated as $n$ .