Together, we make 100

Using SFFT :

Substrating 1 from both sides, this can be factorized as:

$(x - 1) (y + 1) = 99$

Since 99 is equal to: $3^{2} * 11$

$99$ has $(2+1)*(1+1)$ = $6$ divisors that are : $(1,3,9,11,33,99)$

But when $( x - 1) = 99, (y + 1) = 1$ which means that y is $0$ and so we have to exclude it.

So finally $x$ can be $2,4,10,12,34$

And $2 + 4 + 10 + 12 + 34 = 62$

subtracting 1 on both sides we get: $(x-1)(y+1)=99$ Factors of 99 are 1, 3, 9, 11, 33, 99. However $x-1$ can't be 99 since $y$ will be 0. So $x=2, 4, 10, 12, 34$ . Adding these values we get 62

Suppose we find the value of x in terms of y . In doing this, we will find out that x=1+ $\frac{99}{y+1}$

Consider from $\frac{99}{y+1}$

y+1 should be a factor of 99. So these are the possible positive values of y+1; 1,3,9,11,33,99 , which are the factors of 99

but if y+1=1, y=0 so 1 is omitted. So going to the 1st equation for x in terms of y. Just substitute the value of y+1 from the equation we can have the values of x.

Let N be the sum of positive integral value of x.

N=34+12+10+4+2 N=62

$xy+x-y=100 \implies x(y + 1) = 100 + y$ . Let us say that for some $x = k, y$ is a positive integer. This would mean, $k(y+1) = 100 + y \implies ky + k = 100 + y \implies y(k-1) = 100 - k \implies y = \frac{100 - k}{k-1}$ .

Since $y$ is an integer, $100 - k \equiv 0 \space (mod \space k - 1) \implies 100 \equiv k \space (mod \space k - 1)$ . Let $m + 1 = k$ for some $m$ . This means $100 \equiv m + 1 \equiv 1 \space (mod \space m) \implies 99 \equiv 0 \space (mod \space m)$ .

The divisors of $99$ are $1, 3, 9, 11, 33, 99$ . Therefore, all possible values of $k$ are $2, 4, 10, 12, 34, 100$ . But, $100$ does not work. So we exclude it. The sum of the rest is $62$ , which is our answer.

Using Simon's Favorite Factoring Trick, we get xy-y+x-1=99 which means (x-1)(y+1)=99 which is a Diophantine Equation. Notice that x-1 can be 1, 3, 9, 11, or 33, but not 99 because then y=0 which is not positive. So those values of x-1 mentioned above give x=2, 4, 10, 12, or 34, and 2+4+10+12+34=62, our answer.

=(y*(x-1)+x=100) Only x even will be allowed because even + even = even and if x=odd then first term will get even and even + odd can never be even. Then Try and substitute all combo to get x=2,4,10,12,34