Proof that $5x^2 + 4y^3 = 51$ has no solutions for $x,y \in \Bbb Z^+$

Question

So I have been given this question to answer:

Give a proof by cases to show that the equation $5x^2 + 4y^3 = 51$ does not have any solution $x,y \in \mathbb Z^+$

I am assuming that I could give two cases for this problem:

Express everything in terms of y = $((51-5x^2)/4)^{1/3}$
Express everything in terms of x = $((51-4y^3)/5)^{1/2}$

Since $4^{1/3}$ is irrational and $5^{1/2}$ is also irrational, then there are no integers that make this solution true.

Do I have this right?

No. Note that $(5/5)^{1/2}$ is rational. A quotient of two irrational numbers can be rational. — egreg, Oct 20 '16 at 16:09
The equation is almost trivial if $;Z+;$ is the naturals $;\Bbb N;$ ... — DonAntonio, Oct 20 '16 at 16:10
I have posted an answer using modular arithmetic. Please tell me if you have any problem. — Soham, Oct 20 '16 at 16:16

score 5 · Accepted Answer · answered Oct 20 '16 at 16:09

5

You’re making it much more complicated than necessary. If $x$ and $y$ are positive integers such that $5x^2+4y^3=51$, then certainly $4y^3<51$, and a little arithmetic quickly shows that $y$ must be $1$ or $2$: $4\cdot3^3=4\cdot27>51$. Now you have two cases, $y=1$ and $y=2$; substitute those values of $y$, solve for $x$, and verify that $x$ is not an integer.

answered Oct 20 '16 at 16:09

Brian M. Scott

616,228

Thanks! Is it necessary to do the same for x? Or these two cases are sufficient to prove this statement? – Jake0991 Oct 21 '16 at 02:46
1

@Jake0991: You’re welcome. We’ve seen that $y=1$ and $y=2$ are the only possibilities for a solution, so once you show that there is no solution with either of them, you’ve shown that there is no solution at all: you’re done. – Brian M. Scott Oct 21 '16 at 13:56

score 4 · Answer 2 · answered Oct 20 '16 at 16:14

4

$5x^2 \equiv 0,1 \pmod{4}$ (How?)

$4y^3 \equiv0 \pmod {4}$

$\therefore 5x^2+4y^3\equiv0,1\pmod{4}$

But, $51\equiv 3 \pmod{4}$

So, there are no such positive integers (rather, integers!) $x,y$.

answered Oct 20 '16 at 16:14

Soham

9,990

I kind of understand the concept of modular arithmetic (the only place I've used it is in programming), however, I have not yet arrived at that specific chapter yet. – Jake0991 Oct 20 '16 at 16:21
@Jake0991 Do you know that a perfect square number always gives remainder $1$ or $0$ when divided by $4$? – Soham Oct 20 '16 at 16:23
Didn't know this before but now that I think about it yes, it makes sense – Jake0991 Oct 20 '16 at 16:30

score 0 · Answer 3 · answered Oct 20 '16 at 19:14

0

It is immediate because $$5x^2=51-4y^3$$ gives as only possibility $5x^2\in\{47,19\}$

answered Oct 20 '16 at 19:14

Piquito

29,594

Proof that $5x^2 + 4y^3 = 51$ has no solutions for $x,y \in \Bbb Z^+$

3 Answers3