Linear Equations?

Question

Consider the equation 2x + 5y = xy - 1. What is the number of integer pairs (x, y) that solve the given equation?

Answer 1

5y +1 =x(y-2)
the problem boild down to "when y-2 is a divisor of 5y+1"
.
5y+1 cannot be bigger(in absolute value) than 10(y-2), except for those y less than 5. Then you can try the remaining cases:
x:0(no),1(no),2(no),3(no),4(yes, y=-9),5 (no),6(yes, y=13)
7(no),8(no),9(no),10(no)
so i got 2 pairs (4,-9),(6,13)
now try the values of y less than 5,
just treat all this cases and use the fact that i mentioned above, x(y-2) becomes "bigger" than 5y+1, when x is big enough

Answer 2

There are exactly 3 integer solutions:

[x,y] = [-6,1], [4,-9], and [6,13]

The proof is long but in short, this is what you do: First notice that y must be ODD since otherwise 2x+5y is EVEN and xy-1 is ODD. Now that you know that y is ODD, it follows that x must be EVEN. Since if it was not, then xy-1 is EVEN and 2x+5y is ODD as a summ of an even and odd number.

So say x = 2K, y = 2L+1. Then the equation is

4x + 5(2L+1) = 2k(2L+1) - 1

and can be simplified as

L = (K+3)/(2K-5)

So we are looking for integer K such that (K+3) is an integer multiple of (2K-5). Now notice that 2K-5 very quickly outgrows K+3; for K = 9 and bigger [for K=9 we get K+3 = 12 and 2K-5 =13]. So K cannot be bigger than 8. Similar reasoning shows that K cannot be smaller than -3.

Now try all options for K; that is K = -3,-2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8 and backsolve for L. If L is integral; GOOD! You only need to solve for back for x and y. If not, this does not lead to a solution.

All solutions you get are the ones above; originating from
K = -3, 2 and 3; in that order.