Let $\mathbb F$ be field with $16807=7^5$ elements.What is the cardinality of the set given by $\{a^7-b^7:a,b\in \mathbb F\}$?

Question

Let $\mathbb F$ be field with $16807=7^5$ elements.What is the cardinality of the set given by $\{a^7-b^7:a,b\in \mathbb F\}$?

how to approach this problem?

Answer 1

First of all notice, that $a^7-b^7 = a^7+(-b)^7 = (a-b)^7$ for all $a,b \in \mathbb{F}$ according to https://en.wikipedia.org/wiki/Freshman%27s_dream , this means that $$\{ a^7-b^7 \: | \: a,b \in \mathbb{F} \} = \{x^7 | x\in \mathbb{F}\}$$ Now $x\mapsto x^7$ is a ring homomorphism (this is again a consequence of freshmans dream), and if we can prove that it is injective, then we must have $|\{x^7 | x\in \mathbb{F}\}|=|\mathbb{F}|$. Suppose therefore that $a^7=b^7$. Then $a^7-b^7 = (a-b)^7 = 0$, which means that $a=b$ proving injectivity.