Find number of solutions of [ x ] × [ y ] = x + y ?

or:Find number of solutions of [ x ] × [ y ] = x + y ?or:x = 2y = 2or:A bit of thinking will help you realize that {x,y} = {2,2} and {0,0} are

or:Find number of solutions of [ x ] × [ y ] = x + y ?

or:x = 2y = 2

or:A bit of thinking will help you realize that {x,y} = {2,2} and {0,0} are two solutions to the problem, because: 2 + 2 = 2 * 2 = 4 0 + 0 = 0 * 0 = 0 But are there other solutions? To find out, we need to solve the equation using a little algebra: x + y = xy 0 = xy - x - y 0 = (x-1)(y-1) - 1 (Confused? See the Postscript below) 1 = (x-1)(y-1) 1/(x-1) = y-1 y = 1/(x-1) + 1 So, there are an infinite number of real number solutions; to find a given y, just compute y = 1/(x-1)+1 (as long as x is not equal to 1; there's no real number solution in that case). For example, the pair {1.5,3} works, because: 1.5 + 3 = 1.5 * 3 = 4.5 Now, let's narrow the question - are there any other real number solutions where x=y, other than 0 and 2? It turns out the answer is no, and here's why, starting with the equation above and changing it since now x=y: 0 = (x-1)(y-1) - 1 (see above) 0 = (x-1)^2 -1 (because x=y now) 1 = (x-1)^2 +1 or -1 = x-1 x = 0 or 2 Now, instead of limiting ourselves to x=y, let's allow any value of x and y, but only if they are both integers. Given that variation, are there any other integer pair solutions? The short answer is no - if you limit yourself to integers, 0 and 2 are all that's possible. Here's why. Since integers are a subset of the real numbers, the equation given above applies: y = 1/(x-1)+1 So, for y to be an integer, 1/(x-1) has to be an integer. For a fraction to produce an integer when there's a one on top, its denominator has to have an absolute value less than one, so: |x-1|

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