0

I have done a number of congruence questions but then I encountered this question:

$7^{x+2} ≡ 5(mod 29)$

How do you go around solving this? I thought of splitting the powers to $7^x$ and $7^2$ but then I got stuck trying to evaluate it...

2 Answers2

0

Evaluating powers of $7$, mod $29$, we get \begin{align*} 7^0 \equiv 1\;(\text{mod}\;29)\\[4pt] 7^1 \equiv 7\;(\text{mod}\;29)\\[4pt] 7^2 \equiv 20\;(\text{mod}\;29)\\[4pt] 7^3 \equiv 24\;(\text{mod}\;29)\\[4pt] 7^4 \equiv 23\;(\text{mod}\;29)\\[4pt] 7^5 \equiv 16\;(\text{mod}\;29)\\[4pt] 7^6 \equiv 25\;(\text{mod}\;29)\\[4pt] 7^7 \equiv 1\;(\text{mod}\;29)\\[4pt] \end{align*} after which, the cycle repeats forever.

It follows that no power of $7$ is congruent to $5$, mod $29$.

quasi
  • 58,772
0

$2$ can be proved to be a primitive root

$2^6\equiv6,2^7\equiv12,2^8\equiv-5,2^{14}\equiv12^2\equiv-1\pmod{29}$

$5=(-5)(-1)\equiv2^{8+14}$

$7\equiv29+7\equiv6^2\equiv(2^6)^2$

Using discrete logarithm, $(x+2)12\equiv22\pmod{\phi(29)}$

lab bhattacharjee
  • 274,582
  • https://math.stackexchange.com/questions/741832/linear-congruence-theorem-are-these-solutions-too-whered-they-hail-from – lab bhattacharjee Nov 16 '18 at 05:06
  • i.e. $\bmod 29!:\ 5$ is not a $4$th power $(4\nmid 22)$ but $7$ is $(4\mid 12)$ so $7^n$ is too, thus $5\not\equiv 7^n.\ $ Said equivalently $,7^7\equiv 1,$ but $5^7\not\equiv 1\ \ $ – Bill Dubuque Nov 16 '18 at 17:57