The question is:
$$10 \equiv M^5 \mod{35}$$
How do I isolate and solve for $M$?
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3The numbers are so very small, just do it by trial and error. Note: since $\gcd(10,35)=5$ we know that $5,|,M$ so, really, there are very few values to try. – lulu Apr 13 '19 at 21:15
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See this answer. – Bill Dubuque Apr 17 '19 at 18:21
2 Answers
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Hint:
Use the Chinese remainder theorem: $\;\mathbf Z/35\mathbf Z\simeq \mathbf Z/5\mathbf Z\times \mathbf Z/7\mathbf Z$. As the image of $10$ by this isomorphism is the pair $\;(0\bmod 5, 3\bmod 7)$, you first have to solve each congruence equation: $$x^5\equiv 0\mod 5,\quad y^5\equiv 3\mod 7,$$ then put all the possible pairs together with the inverse isomorphism.
Note that $x^5\equiv 0\iff x\equiv 0\mod 5$. As to the congruence modulo $7$, since $y$ is necessarily coprime to $7$, by Fermat's theorem, we have $y^6\equiv 1\mod 7$, so that $y^5\equiv y^{-1}$ and ultimately the solution is $$y\equiv 3^{-1}\equiv 5\mod 7.$$
Can you end the calculations?
Bernard
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Chinese renainder theorem is one way. Another way is using Euler's totient theorem (since the order of all elements divide the order of the group), polynomial remainder theorem and modular reduction. Note by Euler's totient theorem that $$M^{24}\equiv 1\bmod 35\implies M^{25}\equiv M \bmod 35$$ reduce the original congruence to linear equation form: $$M^5=35x+10$$ Note: $$(M^5)^5=M^{25}\equiv M \bmod 35$$ apply to both sides:$$M\equiv(35x+10)^5$$ reduce using polynomial remainder theorem to:$$M\equiv (10^5=100000)\bmod 35$$ aka $$M\equiv 5\bmod 35$$
no modular inverses, or roots required.