 # 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}. tabita57i 2020-11-27 Answered

Using Fermat's Little Theorem, solve the congruence .
Write your answer as a set of congruence classes modulo 107, such as {1,2,3}.

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Step 1
Fermat's Little Theorem says that if p is a prime number and a is an integer such that p does not divides a, then ${a}^{p-1}\equiv 1\text{mod}p$. We need to find the solutions for .
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's Little Theorem, it follows that
.
Taking p = 107, we have

${x}^{108}={x}^{2}\cdot {x}^{106}$

We also have

${x}^{424}={x}^{4\cdot 106}$

Step 2
Now, we can write the given congruence as
.

This given us
Solving this congruence using the basic procedure, we get the solutions x=1, 103. Therefore, the required set of congruence classes is {1, 103}.

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