Solving Diophantine equation without brute force

Question

I am having trouble trying to solve this Diophantine equations: $$ ^3=4^2+4−3 $$

I was wondering if anyone could help me find the integer solutions to these and any advice of techniques to use?

Thank you!

What are your thoughts on the problem? What have you tried and where did you get stuck? — Servaes, Feb 25 '19 at 10:14
I was trying to get it in terms of a number=(...)(...), I got x^3-4=(2y+1)^2 but not sure where to go — , Feb 25 '19 at 10:15
$x^3-4=(2y+1)^2=4y^2+4y+1$ is not the same as $x^3=4y^2+4y-3$. — Dietrich Burde, Feb 25 '19 at 10:26

Servaes · Accepted Answer · 2021-02-05T21:38:31.737

3

The right hand side can be factored as $$x^3=4y^2+4y-3=(2y+1)^2-4=(2y+3)(2y-1).$$ The gcd of the two factors on the right hand side divides their difference, which is $4$, but they are both odd so they are coprime. It follows from unique factorization that both factors are perfect cubes. But no two perfect cubes differ by $4$, a contradiction. Hence no integral solutions exist.

edited Feb 05 '21 at 21:38

answered Feb 25 '19 at 10:15

Servaes

67,306
8
82
171

Then could I say either: x=2y+3 and x^2=2y-1 OR x=2y-1 and x^2=2y+3 ? – Feb 25 '19 at 10:19
$2y+3-(2y-1)=4$ so you can not assume 2y-1 and 2y+3 are coprime. – Word Shallow Feb 25 '19 at 10:22
how would i finish this off? – Feb 25 '19 at 10:35
Thank you very much :) – Feb 25 '19 at 10:39
How did you do the cases for m=1,2? – Feb 27 '19 at 15:47

Dietrich Burde · Answer 2 · 2019-02-25T10:27:39.473

1

The equation in equivalent to $$ x^3+4=(2Y+1)^2=y^2, $$ which is Mordell's equation. This has been solved here:

Solution to Mordell's Equation $y^2=x^3+4$

We see that $2Y+1=\pm 2$ is impossible, hence there is no solution.

edited Feb 25 '19 at 10:27

answered Feb 25 '19 at 10:22

Dietrich Burde

140,055

Word Shallow · Answer 3 · 2019-02-25T13:48:14.347

0

Another way: You can rewrite the equation: $$x^3+8=4y^2+4y+5$$

Or $$(x+2)(x^2-2x+4)=4y^2+4y+5$$

Let $(x+2;x^2-2x+4)=d (d\in N^+)$

Note that : $x^2-2x+4=x(x+2)-4(x+2)+12\Rightarrow d\mid 12$

We will prove $4y^2+4y+5\not\mid12$. Assume $4y^2+4y+5\mid12$, then we have $4y^2+4y+5=12k(k\in Z)$

$\Leftrightarrow (2y+1)^2=12k-4=4(3k-1)$

But $(2y+1)^2\equiv0;1\pmod{3}$ and $4(3k-1)\equiv2\pmod{3}$

Can you continue?

edited Feb 25 '19 at 13:48

answered Feb 25 '19 at 11:00

Word Shallow

2,010
2
13
23

Not sure, quiet new to this, could u explain? – Feb 25 '19 at 12:45
@Jamesodare: where do you feel confused? – Word Shallow Feb 25 '19 at 13:45
1

Your exposition could use a few more words and some more commonly accepted notation to be more comprehensible. I have never seen $\vdots$ to mean divisibility; the standard symbol is $\mid$ which is \mid. – Servaes Feb 25 '19 at 13:46
Sorry about that i'm bad at $\LaTeX$. – Word Shallow Feb 25 '19 at 13:49
@Servaes : Thank you for the inspiring advice. ;) (+1) – Word Shallow Feb 25 '19 at 13:56
1

Oh so contradiction, so 4y^2+4y+5 (not |)12, so there are no int solutions? – Feb 25 '19 at 16:44
Yes, 4y^2+4y+5 (not |)12 but $(x+2)(x^2-2x+4)|12 $ so there are no integer roots. – Word Shallow Feb 25 '19 at 16:51

Solving Diophantine equation without brute force

3 Answers3