Slowly Solve Diophantine Equation x²(y-1)+y²(x-1)=1
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I have seen on YouTube how the equation is solved with the
substitution:
 1) u=x+1
 2) v=y+1

I want to solve it without substitution and to show some thinking.

So, the requested solutions are pairs of integers (x,y) satisfying the equation.
Let us start by expanding:

   xy(x + y) - x² - y² = 1

==> xy(x+y) - (x² + y²) = 1

Now, a good way to complete x² + y²  to a multiple of x + y is to add 2xy
to it. To be more precise, we need to subtract it  and add it:

   xy(x + y) + 2xy - (x² + y²) - 2xy ==>

==> xy(x + y) + 2xy - (x²+ 2xy + y²) = 1

==> xy(x + y + 2) - (x + y)(x + y) = 1

Great! Now, let us complete (x+y)(x+y) to a multiple of (x + y + 2):

   xy(x + y + 2) - (x + y + 2 - 2)(x + y) = 1 ==>

==> xy(x + y + 2) - (x + y + 2)(x + y) + 2(x + y) = 1 ==>

==> (xy - x - y)(x + y + 2) + 2(x + y) = 1 ==>

==> (xy - x - y)(x + y + 2) + 2(x + y + 2 - 2) = 1 ==>

==> (xy - x - y)(x + y + 2) + 2(x + y + 2) - 4 = 1 ==>

==> (xy - x - y + 2)(x + y + 2) = 5

Great! Now the divisor of 5 are 1, -1, 5 and -5, and each of the factors in
parentheses is an integer. So let us solve systems of equations:

System 1:
   x + y + 2 = 1 ==> x + y = -1
   xy - x - y + 2 = 5 ==> xy + 3 = 5 ==> xy = 2

   Plug y = -1 - x = - (x + 1)  into the second equation:

   x(x + 1) = -2 ==>

==> x²+ x + 2 = 0

The discriminanat is 1-4(2)(1)=1-8=-7<0
Thus, there are no real solutions here, let alone integer solutions.

System 2:
--------
   x + y + 2 = -1 ==> x + y = -3
   xy - x - y + 2 = -5 ==> xy + 3 + 2 = - 5 ==> xy = -10

   The solutions are:
     (x, y) = (2, -5) and (x,y) = (-5, 2)

System 3:
--------
   x + y + 2 = 5 ==> x + y = 3
   xy - x - y + 2 = 1 ==> xy - 3 + 2 = 1 ==> xy = 2

   The solutions are:
     (x, y) = (1, 2) and (x, y) = (2, 1)

System 4:
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   x + y + 2 = -5 ==> x + y = -7
   xy - x - y + 2 = -1 ==> xy + 7 + 2 = -1 ==> xy = -10

   Plug y= -7 - x into the second equation:

   x(-7 - x) = -10 ==> x(x + 7) = 10 ==>

==> x² + 7x - 10 = 0

The discriminant is 7² - 4(-10) = 89. It is not a perfect square, so there are
no rational solutions here, let alone integer solutions.

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Thus, the solution set for (x,y) is:
 {(2, -5),(-5, 2), (1, 2), (2, 1)}