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So I'm using the fact that 2 is a primitive root modulo 53, I'm solving $x^5 \equiv 8\mod{53}$

Originally I was trying to rewrite both sides in terms of two so had the following:

let $x=2^y$ for y between 1 and 52... $(2^y)^5 \equiv 8\mod{53}$,
$2^{5y} \equiv 2^3\mod{53}$,
$5y \equiv 3\mod{53}$
which I believe has a solution of $y=32$ so for $x^5 \equiv (2^{32})^5 \equiv 8 \mod{53}$.

I thought there would be multiple solutions so looked at a different source following a slightly different method:

again let $x=2^y$ for y between 1 and 52...
$(2^y)^5 =8 \mod{53} $
$2^{5y} =8=2^3$
$2^{5y-3}=1$ as 2 is a primitive root, the order is 52 and therefore 52 must divide $5y-3$, so $y=11$ and $x=2^11$ so $x^5 \equiv (2^{11})^5 \equiv 8\mod{53}$.

If there are any errors in either method please let me know as I'm assuming it is my fault and these should result in the same answer? Any advice on which is correct and how to find other solutions would be really appreciated

amWhy
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thomasmaths
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  • If $n=11$, $5n\equiv 3 \pmod {52}$, by inspection. Thus the answer you want is $2^{11}\equiv 34 \pmod {53}$. Since $5,\nmid ,52$, there is only one solution. – lulu Dec 15 '21 at 18:10
  • Easier: by here$\bmod 53!:\ x\equiv 2^{\large\color{#c00}{3/5}}\equiv 2^{\large\color{#c00}{11}}\equiv 34,,$ by $\bmod \color{#0a0}{52}:\ \color{#c00}{\dfrac{3}5}\equiv \dfrac{55}5\equiv \color{#c00}{11},,$ which doesn't require knowing any primitive root. – Bill Dubuque Dec 15 '21 at 18:33
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    The error in your first method is it should be, as above $,\color{#c00}5y\equiv \color{#c00}3\pmod{\color{#0a0}{52}},,$ not $\bmod 53\ \ $ – Bill Dubuque Dec 15 '21 at 18:37
  • Thanks for your help, I'm looking at congruences using the same primitive root modulo 53, so can I just ask if this would be different if the congruence contained a prime number e.g. 11mod53 for example? – thomasmaths Dec 15 '21 at 20:21

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