Using Fermat's Little Theorem, solve the congruence 2⋅x425+4⋅x108−3⋅x2+x−4≡0 mod 107. Write your answer as a set of congruence classes modulo 107, such as {1,2,3}.

Question

Using Fermat&#039;s Little Theorem, solve the congruence 2⋅x425+4⋅x108−3⋅x2+x−4≡0 mod 107. Write your answer as a set of congruence classes modulo 107, such as {1,2,3}.

Jaylen Fountain · Accepted Answer

Step 1According to Fermat&#039;s Little Theorem, if an is an integer such that p does not divide it and p is a prime number, then&amp;nbsp;ap−1≡1modp. We need to find the solutions for&amp;nbsp;2·x425+4·x108−3·x2+x−4≡0&amp;nbsp;mod&amp;nbsp;107.It can be seen clearly that 107 can not be the answer to the given congruence, as otherwise, we will have 107 lambda−4 on the left side which is clearly not 0 mod 107. Therefore, from Fermat&#039;s Little Theorem, it follows thatxp−1=x106≡1&amp;nbsp;mod&amp;nbsp;107.Taking p = 107, we havex106≡1&amp;nbsp;mod&amp;nbsp;107x108=x2⋅x106⇒x108≡x2&amp;nbsp;mod&amp;nbsp;107We also havex106≡1&amp;nbsp;mod&amp;nbsp;107x424=x4⋅106⇒x424≡1&amp;nbsp;mod&amp;nbsp;107⇒x425≡x&amp;nbsp;mod&amp;nbsp;107Step 2Now, we can write the given congruence as2·x+4·x2−3·x2+x−4≡0&amp;nbsp;mod&amp;nbsp;107.⇒x2+3x−4≡0&amp;nbsp;mod&amp;nbsp;107This given us&amp;nbsp;(x−1)(x+4)≡0&amp;nbsp;mod&amp;nbsp;107Solving this congruence using the basic procedure, we get the solutions x=1, 103. Therefore, the required set of congruence classes is {1, 103}.

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Using Fermat's Little Theorem, solve the congruence 2 * x^(425) + 4 * x^(108) - 3 * x^2 + x-4 -=0 mod 107.Write your answer as a set of congruence classes modulo 107, such as {1,2,3}.

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