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Exercise 8.4.1 from What is Mathematics? By Richard Courant

Find pairs $(x, y)$ for $\{x, y\} \in \mathbb{Z} $, that satisfy $153y^3 = 5 - x^3 $

 #1
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Hi Maths solver,

 

I have played with it but I have no idea.   

Is the worked answer given?

 Aug 6, 2021
 #2
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@Melody

 

Note $153 = 3^2 \cdot 17$

Rearrange the orig. equation: $x^3 = 5 - 153y^3$

You try all the possible combinations of possible modulos.

At one point or another, you end up with $\pmod 9$

So, if we try to simplify mod 9...

$x^3 \equiv 5 \pmod 9$

Consider this.

You have $x = n-1, n, n+1, n = 3k, k \in \mathbb{Z}$

If you have $x=n,$ obviously $x^3 \equiv 0 \pmod 9$

But…

If you have $x=n-1,$ you have $n^3 - 3n^2 + 3n - 1 \equiv -1 \pmod 9$

If you have $x = n+1,$ you have $n^3 + 3n^2 - 3n + 1 \equiv 1 \pmod 9$

So $x^3 \equiv \{-1, 0, 1\} \pmod 9$

But we have $x^3 \equiv 5 \pmod 9$ which is a contradiction. Thus there are 0 solutions.


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