Show that if $x$ is an integer, then $x^{10} \equiv \{-1, 0 ,1\} \pmod{25}$

Question

Show that if $x$ is an integer, then $x^{10} \equiv \{-1, 0 ,1\} \pmod{25}$

Well, the first case if $(25, x)$ is not 1, so $x$ is a multiple of $5$ and $x^{10} \equiv 0$.

If $(25, x) =1$, $x^{20} \equiv 1$ by Euler's Theorem.

I am just learning number theory, how can we jump from that to $x^{10} \equiv 1$ or $-1$?

That inference is proved in the linked dupes more generally for modulus $p^2$ for odd prime $,p\ \ $ — Bill Dubuque, Apr 12 '22 at 01:37

Anatoly · Answer 1 · 2022-04-12T06:44:42.297

2

Hint: note that, for $x$ integer, $$x^2=\{1,0,-1\} \pmod 5$$

edited Apr 12 '22 at 06:44

answered Apr 11 '22 at 20:14

Anatoly

17,079

Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Apr 12 '22 at 01:37

score 1 · Accepted Answer · answered Apr 11 '22 at 20:10

1

$x^{20} \equiv 1 \pmod{25}$ implies $25 \mid (x^{10}-1)(x^{10}+1)$. Try to work from that.

answered Apr 11 '22 at 20:10

radekzak

1,831

Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Apr 12 '22 at 01:37

score 0 · Answer 3 · answered Apr 11 '22 at 22:29

0

If $d=(x,25)=1$, then by Euler's Theorem we have $$x^{20}\equiv 1\pmod{ 25}\implies x^{10}\equiv \pm 1\pmod 5$$ That would mean $x^{10}\pm 1\equiv 0\pmod 5$ Hence$$ (x^{10}\pm 1)^2=x^{20}\pm 2x^{10}+1\equiv 2\pm 2\equiv 0 \text{ or }-1\pmod {25}$$

Now if $d\ne 1$ then $x\equiv 0\pmod 5\implies x^2\equiv 0\pmod{25}\implies x^{10}\equiv 0\pmod{25}$.

answered Apr 11 '22 at 22:29

PNT

4,164

Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Apr 12 '22 at 01:38

Show that if $x$ is an integer, then $x^{10} \equiv \{-1, 0 ,1\} \pmod{25}$

3 Answers3