Let $a$ be an integer and $p$ an odd prime number. Suppose $\bar{a}$ has order $h>1$ in $\mathbb{Z}_p^*$, how can we show that $$a^{h-1} + a^{h-2} + ... + a + 1\equiv 0\mod p?$$
Asked
Active
Viewed 26 times
-4
-
1I edited your question. Learn to use mathjax if you are planning to ask more questions on this site. People don't like badly asked and poorly edited questions. – Mathematician 42 Mar 18 '17 at 13:25
-
Is h an integer or just any real number greater than 1? – Michael R. Chernick Mar 18 '17 at 13:42
-
@MichaelChernick h is order of a modulo p , then h must be the positive integer – user426865 Mar 18 '17 at 13:45
-
Okay I can buy that. – Michael R. Chernick Mar 18 '17 at 13:46
-
This is true only if $a\not\equiv 1\mod p$, unless $h=p$. – Bernard Mar 18 '17 at 13:50
2 Answers
2
Use the high school factorisation $$x^n-1=(x-1)(x^{n-1}+x^{n-2}+\dots+x+1).$$ Note the relation is valid only if $a\not\equiv 1\mod p$.
Bernard
- 175,478
-
I don't get why this was downvoted. Ok I have said the same a bit earlier, but it doesn't deserve a downvote for that. +1 to compensate. – Mathematician 42 Mar 18 '17 at 13:22
-
1@Mathematician 42: I was writing my answer while you were posting, apparently. I left it because I added a remark on the validity of the relation in the O.P.'s question, but you may as well add it to your own answer, and I'll delete mine. – Bernard Mar 18 '17 at 13:28
-
1The validity is ok, the order of $a$ is assumed to be different from one. – Mathematician 42 Mar 18 '17 at 13:32
1
Hint : You have that $$(a-1)(a^{h-1}+a^{h-2}+\dots +1)=a^h-1.$$ Hence $(a-1)(a^{h-1}+a^{h-2}+\dots +1)\mod p=0$.
Mathematician 42
- 12,591