Find out the answer to "Solvingx5−2x3y2+xy4=y2"

Bruce Rosario

Answered question

2022-04-21

Solving

x^{5} - 2 x^{3} y^{2} + x y^{4} = y^{2}

Ann Mathis

Beginner2022-04-22Added 11 answers

x^{5} - 2 x^{3} y^{2} + x y^{4} = x {(x^{2} - y^{2})}^{2} = y^{2}

As pointed out in the comments,

x^{2} - y^{2}

must be a factor of y. Hence both

x \pm y

are factors of y. Write

y = m (x - y), y = n (x - y)

where m, n are integers. Then if

y \neq 0

n y = m n (x - y), m y = m n (x + y)

(m - n) y = 2 m n y \leq a d s \to - 1 = 4 m n - 2 m + 2 n - 1 = (2 m + 1) (2 n - 1)

Thus

{2 m + 1, 2 n - 1} = {\pm 1}, i . e . m = n = 0 or m = - 1, n = 1

.
The first possibility gives

y = 0

, which leads to

x = 0

.
The second possibility gives

x = 0

, which leads to

y = 0

.
Thus (0,0) is the only solution.

ophelialee4xn

Beginner2022-04-23Added 14 answers

I use Ben's hint. Clearly,

x = 0 \Leftrightarrow y = 0

. Henceforth, x and y are nonzero. Since x is a square, we write

x = t^{2}

and get

t {(t^{4} - y^{2})} = y .

Thus, y|y and, therefore

t^{2} ∣ y

. Writing

y = t^{2} u

and cancelling

t^{2}

, we get

u = t^{3} (1 - u^{2}),

which is impossible because of at least two contradictions (using that u is nonzero):
(1)

| u | + 1

divides |u|;
(2)

t^{3}, |, u, |, t^{3}

because u and

u^{2} - 1

are co'.

(1 - u^{2}

would have to be

\pm 1

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