# How can I prove that if $a^7 = b^7$ then $a=b$, with $a,b \in \mathbb{Z}$

I've tried with divisibility, meaning that since $a$ divides $a^7$, then $a$ divides $b^7$ and in the same way b divides $a^7$, but I can't seem to go further than this. What properties of the integers could I use to solve this?

• Unique prime factorisation is one option. Noting that $n \mapsto n^7$ is strictly increasing is another. – Daniel Fischer Feb 24 '16 at 23:07

Hint:$$a^7-b^7=(a-b)(a^6+a^5b+...+b^6)=0$$ $$a^6+a^5b+...+b^6$$ is always.....

• Why downvote??? – Domenico Vuono Feb 24 '16 at 23:38

Suppose $a>b\ge0$ and set $a=b+c$, with $c>0$. Then $$a^7-b^7=(b+c)^7-b^7=\sum_{k=1}^7\binom{7}{k}b^{7-k}c^k>0$$ Therefore $a^7>b^7$.

If $a>0\ge b$, then clearly $a^7>b^7$.

If $0\ge a>b$, then $0\le -a<-b$, so by the first step we get $(-b)^7>(-a)^7$ that implies $a^7>b^7$.

So we have proved that, in any case, $a>b$ implies $a^7>b^7$. Of course $a<b$ implies $a^7<b^7$ and, putting these together, we have that $a\ne b$ implies $a^7\ne b^7$.

You can also do it with factorization. Suppose $a^7=b^7$, with $a>0$ and $b>0$ (since $7$ is odd, it is not restrictive).

Let $p$ be a prime dividing $a$ and let $r$ be the maximum exponent such that $p^r$ divides $a$. Then $p^{7r}$ also divides $b^7$. In particular $p$ divides $b^7$, hence it divides $b$. Let $s$ be the maximum exponent such that $p^s$ divides $b$. Then $p^{7s}$ divides $b^7$, so $7r\le 7s$ and $r\le s$. By reversing the roles of $a$ and $b$, we get that $s\le r$.

Hint: $a=b$ iff $v_p(a)=v_p(b)$ for all primes $p$

$v_p(x)$ is the exponent of $p$ in the factorization of $x$ so that $$x=\prod_p p^{v_p(x)}$$

• What does $vp(a)$ means? – Ben-ZT Feb 24 '16 at 23:09
• @Ben-ZT The exponent that has $p$ in the factorization of $a$. – ajotatxe Feb 24 '16 at 23:14

If $|a|>|b|$ then $|a^7|>|b^7|$. Similarly if $|a|<|b|$. So $|a|=|b|$. Since $a^7=b^7$ and $7$ is odd, $a$ and $b$ have the same sign.

$a^7-b^7=(a-b)a^6p(\frac ba)=0$ where $p(x)=x^6+x^5+x^4+x^3+x^2+x+1$ is the cyclotomic polynomial of degree $6=7-1$ and all of these polynomials with degree $p-1$ where $p$ prime it is well known to be irreducible so $a^6p(\frac ba)\ne 0$ since $(a,b)\in \mathbb Z^2$. Hence $a-b=0$

If $a^7 = b^7$, then $0 =a^7-b^7 = a^7(1-(b/a)^7)$.

Consider $f(x) = 1-x^7$. $f(1) = 0$. $f'(x) =-7x^6 \le 0$ since $x^6 \ge 0$ with equality only if $x = 0$.

Therefore $f(x)$ can have at most one real root since it is decreasing everywhere except zero where it is zero.

Therefore $1-x^7$ can have at most one real root. Since $f(1) = 0$, that is its only real root.

This works for $a^{2n+1}-b^{2n+1}$ for any positive integer $n$.

• Because I wanted to do it this way. – marty cohen Feb 24 '16 at 23:25