+876 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 $
+118687 Hi Maths solver,
I have played with it but I have no idea.
Is the worked answer given?
+876 @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.
