One Simple Trick Solves This Equation Instantly

[daxiang]

If x >= y >= 2, then (x+y)^2 <= (2x)^2 <= (xy)^2 < 2(xy)^2, the equation does not hold, thus either x or y must be 1, and the other value 2 is easily obtained

[spiderjerusalem]

x+y∈odd
(since there is the term 2(xy)², which is even)

(x+y)²–1 = 2x²y²
(x+y+1)(x+y–1) = 2x²y²
x+y–1 = 2 ⇔ x+y = 3 ⇒ (2, 1) or (1, 2)
which also satisfies the other equality (that is (x+y+1)=(x²y²))

[matniet]

Could the Pell's equation method be used here?

[justinnitoi]

Let a = x+y and b = xy
2b^2 = (a-1)(a+1)

Notice that a-1 and a+1 must be even so let a=2k+1
2b^2 = (2k)(2k+2)
But since there is at least 2 factors of 2 in the RHS, b must have at least 1 factor of 2 so let b = 2c
8c^2 = (2k)(2k+2)
2c^2 = k(k+1)

Notice that either k is even or k+1 is even.
If k is even:
k = 2d
k+1 = c^2/d
So k+1 - k = c^2/d - 2d = 1
d(c^2/d^2 - 2) = 1
So d = 1 and c^2 = 3
No solution.

If k+1 is even:
k+1 = 2d
k = c^2/d
k+1 - k = 2d - c^2/d = 1
d(2-c^2/d^2) = 1
So d = 1 and c^2 = 1 so c = 1
Then k = 1, b = 2
a = 3

x + y = 3 and xy = 2
y = 3-x

x^2 - 3x + 2 = 0
x = 1, 2
y = 2, 1

Therefore, (x, y) = (1, 2), (2, 1)

[noodlesodyssey]

You can use simple modular arithmetic for this question as well. Note taking mod x or mod y leaves x == +-1 mod y, or vice versa. The symmetry of the equation means the other variable is of the same form. If abs(x - y) >= 2 then this is only possible W.O.LO.G for one to = 1, which leads to the only solution. You can easily check when they are two consecutive integers that it leads to an unsolvable cubic in natural numbers.

[tianqilong]

my method is kinda simpler:

First to prove that: at least one of x and y needs to be 1. Because if x and y both are not 1, then x + y <= xy, equals only when x = y = 2, which does not fit for the question. So just let x =1, and solve a quadratic equation. the answer been: x = 1, y = 2 as the only valid sol.

[wesleydeng]

Divide both sides of the equation by x^2*y^2 one gets: (1/x+1/y)^2 = 2 + 1/(x^2y^2). If x and y>=2, LHS <= 1. So one of x, y must be 1.

[yoav]

x+y>xy, so 1/x+1/y>1 so x(or y from symmetry) is 1. So we get 1+2y-y^2=1 y(2-y)=0. Sk the sol (1, 2) and (2, 1) from symmetry.

[germanbzhitskikh]

Why we can't do
x+y-Sqrt(2)xy=1, x+y+Sqrt(2)xy=1
or
x+y-Sqrt(2)xy=-1, x+y+Sqrt(2)xy=-1
and solve systems?

[satyajitrout]

Solve using Pell's equation ..
Pell's equation x^2-2y^2=1 According to Catalan's conjecture, (3, 2) is a solution of the equation .
Our question is (x+y)^2-2(xy)^2=1
So x+y=3 and xy=2 it is clear that x is either 2 or 1 similarly y can be 2 or 1
So solution of the equation is {(1, 2), (2, 1)}

[ابوسالم-هم]

Thanks, you give us an excellent idea.

If you solve as deference between two square
And set as diphont equation, so will come easier than this solution

[oguzhanozdogan]

x+y=m, xy=n
m^2-2n^2=1
Pell equations.

[MyOneFiftiethOfADollar]

“One simple trick solves this equation instantly”

“ Let’s think Critically”

What’s wrong with this picture?

[renesperb]

I found it very easy to find x=1, y=2 or x=2, y=1, just by trying.

[nguyenhongan]

I have this problem, wanna share with u: Find all natural numbers x, n, k that satisfy the following propositions:
1/
2/ n=k+x^(k+x^(k+x^(k+x^n))))