A Nice and Symmetric Equation | Polish Mathematical Olympiad Second Round

[johnvandenberg]

Note: for any quadratic equation z^2 + bz + a, the product of its roots = a, while the sum of its roots = -b.

[Bruno_Haible]

This is a special case of a more general technique/trick: If a problem/equation is symmetric in x and y, rewrite it in terms of a = x+y and b = xy. If a problem/equation is symmetric in x, y, z, rewrite it in terms of a = x+y+z, b = xy+yz+xz, c = xyz. And so on.

[taharanjbar]

If take x'=x-1, y'=y-1.then rewrite the equation becomes
(4+x'+y')(1+x'y')=5
Each parantheses may equal to +1, -1, +5, -5.
Remaining is the same.

[E.h.a.b]

For integer numbers as solution we need one of these conditions (C1), (C2) to be met.
First condition (C1) a: x^2 (y-1) = 1 AND b: y^2 (x-1) = 0
from (C1.a) x^2 (y-1) = 1 we get
y = 1 + 1/x^2
For integer x & y we must have x^2=1 so x= +/-1 and y = 2
substitute in (C1.b) y^2 (x-1) = 0 we get
x = -1 : (2)^2 (-1-1) = - 8 <> 0 refused
x = 1 : (2)^2 (1-1) = 0 accepted so (1, 2) [Answer]

Second condition (C2) a: x^2 (y-1) < 0 AND b: y^2 (x-1) > 1
from (C2.a) x^2 (y-1) < 0 since x^2 > 0 we get
(y-1) < 0 so y < 1
from (C2.b) y^2 (x-1) > 1 we get
x > 1 + 1/y^2 so x > 1
so our 2nd. condition now become (y < 1 and x > 1)
we need to get ride of y^2 (x-1) or x^2 (y-1) to simplify our equation
so we substitute x=2 (x>1) in our equation to get
2^2 (y-1) + y^2 (2-1) = 1
4 y - 4 + y^2 = 1
y^2 + 4 y - 5 = 0
(y-1)(y+5) = 0 and this give us
y = 1 refused since we need y < 1
y = -5 accepted so (2, -5) [Answer]
so our final solution is (1, 2), (2, -5) or (2, 1), (-5, 2)

[noodlesodyssey]

Let p = x -1, q = y -1, then x^2 = (p + 1)^2 and y^2 = (q + 1)^2.
Expanding and simplifying you can factorise qp^2 + pq^2 + 4pq + p + q into:
(pq + 1)(p + q + 4) - 4 = 1 = 5
Solving these individually you get the parentheses to equal +- 1 and + - 5 and testing each case you arrive at the conclusion:
x, y = 1, 2 and -5, 2 and permutations.

[alexey.c]

We could use undetermined coefficients: (xy - x - y + a)(x + y + b) = x²y - x² + y²x - y² + c

[recepduzenli]

Hocam geometri videoları da atarmisiniz kanala.geometria problems

[sea]

Simpler solution:
x and y are relatively prime.
Then by parity if x and y are both different from 0, 1, 2 there is no solution (because the expression would be even)

We finally only need to check x=0, 1, 2 and we get the solutions.

[robertsandy]

Can you please increase your sound levels. Thanks