The product of two numbers X and Y is 3 less than the sum of the numbers. How many such integer pairs of X and Y are possible.

First we need to write the problem in mathematical language i.e. we need to write an equation.

Product of two numbers `xy,` sum of two numbers `x+y` and since the product is less by 3, we have

`xy=x+y-3`

Write `y` in terms of `x.`

`xy-y=x-3`

Factor the left side.

`y(x-1)=x-3`

Check whether `x=1` gives an integer solution.

`0=-2` 

`x=1` does not gives us an integer solution so we can assume that `x ne1` which allows us to divide by `x-1.`

`y=(x-3)/(x-1)`

Now we know...

First we need to write the problem in mathematical language i.e. we need to write an equation.

Product of two numbers `xy,` sum of two numbers `x+y` and since the product is less by 3, we have

`xy=x+y-3`

Write `y` in terms of `x.`

`xy-y=x-3`

Factor the left side.

`y(x-1)=x-3`

Check whether `x=1` gives an integer solution.

`0=-2` 

`x=1` does not gives us an integer solution so we can assume that `x ne1` which allows us to divide by `x-1.`

`y=(x-3)/(x-1)`

Now we know that in order to get an integer solution `x-3`  must be divisible by `x-1,`  bearing in mind that `x` is an integer as well. The only thing remaining to do is to check whether those two numbers are divisible for some values of `x.`  We will start with values of `x` such that `x-1=-1` or `x-1=1` because those are certainly integers.

`x=0`

`y=(-3)/(-1)=3`

`x=-1`

`y=(-4)/(-2)=2`

`x=-2`

`y=(-5)/(-3)!inZZ`

`x=-2` is not a solution. If we keep trying with the smaller and smaller numbers we will get fractions that get closer and closer to 1, but we will never get an integer.

Let us try greater numbers. We already know that `x=1` is not a solution so we will skip that.

`x=2`

`y=-1/1=-1`

`x=3`

`y=0/2=0`

`x=4`

`y=1/3!inZZ`

Again, we see that `x=4` is not a solution and nor is any integer greater that 4.

We can conclude there are 4 such integer pair numbers and they are

`(-1,2), (0,3), (2,-1)` and `(3,0).`