Let (1) the initial equation. (1) (x+y)^2 -2xy -xy(x+y)+1 =0 (2). x+y=s , xy=p with s, p integers as sum and product of integers x, y . (2) : s^2 -2p -ps+1 =0 (3) . The discriminant if (3) it is necessary a perfect square because the (3) have integers roots. Thus (-p)^2 - 4(-2p+1) = k^2 (p+4)^2 - k^2 =20 (p+4+k)(p+4-k) = 20 Factors of 20 : 1, 2, 4, 5, 10, 20 and -1, -2, -4, -5, -10, -20 .. We solve some systems, p+4+k=1 and p+4-k=20 etc .. 😂 Accepted only integer solutions. . .... Another most sophisticated solution. (1) x^2 •(y-1) +y^2 •(x-1) =1 (2) . Let x=r+1 and y=t+1 and (2) : (r+1)^2 •t + (t+1)^2 •r =1 after some algebra . . (rt+1)(r+t+4) =5 and same method as higher up. Solutions: (1, 2), (-5, 2), (2, 1), (2, -5) (symmetry).
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Understood this problem
Let (1) the initial equation.
(1) (x+y)^2 -2xy -xy(x+y)+1 =0 (2).
x+y=s , xy=p with s, p integers as sum and product of integers x, y .
(2) : s^2 -2p -ps+1 =0 (3) .
The discriminant if (3) it is necessary a perfect square because the (3) have integers roots.
Thus (-p)^2 - 4(-2p+1) = k^2
(p+4)^2 - k^2 =20
(p+4+k)(p+4-k) = 20
Factors of 20 : 1, 2, 4, 5, 10, 20 and
-1, -2, -4, -5, -10, -20 ..
We solve some systems,
p+4+k=1 and p+4-k=20 etc .. 😂
Accepted only integer solutions. .
....
Another most sophisticated solution.
(1) x^2 •(y-1) +y^2 •(x-1) =1 (2) .
Let x=r+1 and y=t+1 and
(2) : (r+1)^2 •t + (t+1)^2 •r =1 after some algebra . .
(rt+1)(r+t+4) =5 and same method as higher up.
Solutions:
(1, 2), (-5, 2), (2, 1), (2, -5) (symmetry).
(x,y)=(1,2); (2,1)
(x,y) = (2,1), (1,2).
X=y, x^3 (x-1)=1
Dos casos más -1×-5 =- 5 × -1= 5
xyxy4/xyxyxy^6=xyxyxyxyxy1.2 (xyxyxyxyxy ➖ 2xyxyxyxyxy+1)