Proving $ a^4 \equiv 1 \pmod d$

Question

I need to prove the following statements:

Prove the following statements:

(a) if $a$ is odd then $a^4 ≡ 1 \pmod 4$,

(b) if $5$ does not divide a, then $a^4 \equiv 1 \pmod 5$.

Can I do this inductively? Or should I be adopting another approach? I know for (a), if $a$ is odd, $a^4$ will also be odd, as the product of odd numbers is always odd. This would mean that $4 \mid (a^4 - 1)$, which would always be an even number, but of course not all even numbers are divisible by 4 (or rather $(a^4 - 1)$ would always be an even number). I also know you can obtain all the congruence cases of $1 \pmod 5$ by adding or subtracting $5$, beginning at $1$.

For (a), see http://math.stackexchange.com/questions/1827165/prove-that-for-all-odd-integer-n-n4-1-pmod-16. — lhf, Jul 05 '16 at 14:47

score 3 · Answer 1 · edited Jul 05 '16 at 14:47

3

Hint: For the first $(2n+1)^4=1+4m$

For the second, compute $1^4,2^4,3^4,4^4 \pmod 5$

edited Jul 05 '16 at 14:47

awllower

16,536

answered Jul 05 '16 at 14:44

Tsemo Aristide

87,475

score 2 · Answer 2 · answered Jul 05 '16 at 14:47

2

Since the moduli are small, you can simply go through all the cases:

1. If $a$ is odd, then $a \equiv 1$ or $3 \pmod 4$. $$1^4 = 1 \pmod 4$$ $$3^4 = \ldots \equiv 1 \pmod 4$$

2. If $5$ does not divide $a$, then $a \equiv 1,2,3$ or $4 \pmod 5$. $$1^4 = 1 \pmod 5$$ $$2^4 = \ldots \equiv 1 \pmod 5$$ $$\text{etc.}$$

answered Jul 05 '16 at 14:47

Théophile

24,627

I was thinking of just demonstrating cases for a, is that usually sufficient to prove the claim? – Chris T Jul 05 '16 at 14:49
@ChrisT Yes, and that is the gist of my answer. It is sufficient, so long as you cover all the cases. – Théophile Jul 05 '16 at 14:55

awllower · Accepted Answer · 2016-07-05T14:52:37.263

1

Hints:
a) If $a$ is odd, then $a$ is of the form $2k+1.$ So expand $a^2.$
b) If $5\not\mid a,$ then $a$ is congruent modulo $5$ to one of $1,2,3,4,$ so you can enumerate these four cases.
Also, have a look at Euler's theorem, or just Fermat's little theorem.

Hope this helps.

edited Jul 05 '16 at 14:52

answered Jul 05 '16 at 14:45

awllower

16,536

1

Ah, I think I see it now. 4k^2 + 4k + 1 is obviously divisible by 4 (after you expand out a^2), right? – Chris T Jul 05 '16 at 14:56
Yes, this is what I meant to say. :) – awllower Jul 05 '16 at 14:58
@ChrisT $4k^2+4k+1$ is obviously not divisible by $4$.....$(4k^2+4k+1)-1$ is divisible... – Soham Jul 05 '16 at 16:07
@tatan Thanks for pointing this slip of eyes out. :) BTW, why do you tag $4k^2+4k+1?$ It isn't a user's name, is it? – awllower Jul 05 '16 at 16:08
1

@awllower It's my slip... – Soham Jul 05 '16 at 16:11

score 1 · Answer 4 · answered Jul 05 '16 at 15:02

If $a$ is odd, then $a^4\equiv1\pmod4$:

$a\equiv\color\red1\pmod4 \implies a^4\equiv\color\red1^4\equiv1\pmod4$
$a\equiv\color\red3\pmod4 \implies a^4\equiv\color\red3^4\equiv81\equiv1\pmod4$

If $5$ does not divide $a$, then $a^4\equiv1\pmod5$:

$a\equiv\color\red1\pmod5 \implies a^4\equiv\color\red1^4\equiv1\pmod5$
$a\equiv\color\red2\pmod5 \implies a^4\equiv\color\red2^4\equiv16\equiv1\pmod5$
$a\equiv\color\red3\pmod5 \implies a^4\equiv\color\red3^4\equiv81\equiv1\pmod5$
$a\equiv\color\red4\pmod5 \implies a^4\equiv\color\red4^4\equiv256\equiv1\pmod5$

Proving $ a^4 \equiv 1 \pmod d$

4 Answers4