# Solving an equation and Euler's totient function [closed]

I stumbled upon two exercises in a book I can not seem to solve. Hints and such very much appreciated.

The first is: Has the following a solution $\left( x,y \right) \in \mathbb{Z}$: $$2x^2 + 3y^2 = 24122017$$

The second is: For $n \in \mathbb{N}$ define $\psi(n)=\left\vert \left\{ a \in \left( \mathbb{Z}/n\mathbb{Z}\right)^* \left\vert a^{n-1}\neq 1 \right.\right\vert\right\}$, show that it holds: if $\psi(n)\geq 1 \Rightarrow \psi(n)\geq \dfrac{1}{2}\phi(n)$, where $\phi$ is Euler's totient function.

Edit: For whatever reason I already asked the second question and answered it myself?! Can not remember that though, here is the link for anyone interested. Euler's Totient Function and a Subset of Z/nZ

## closed as off-topic by MatheinBoulomenos, Namaste, Sahiba Arora, Clarinetist, rtybaseJan 10 '18 at 16:15

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If we take mod 3 we get $$2x^2\equiv 2x^2 + 3y^2 \equiv 24122017 \equiv 1$$
so $$2x^2\equiv 1$$
If we multipy this with 2 we get $$4x^2 \equiv x^2 \equiv 2$$ and thus it has no solution.
• Why does it suffice to show this just for mod $3$? – Abbraxas Jan 9 '18 at 16:56
• @Abbraxas - if the original equation had a solution, that solution would also generate a solution of the mod 3 equivalence. But $2 \equiv -1 \mod 3$, so $-x^2 \equiv 2x^2 \equiv 1 \mod 3$ and thus $x^2 \equiv -1 \mod 3$. John Watson just did the same thing without involving negatives. It is a matter of choice which you prefer. Since this equivalence has no solution, the original equation cannot have one either. – Paul Sinclair Jan 10 '18 at 2:47