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# I need help

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Find all residues \$a\$ such that \$a\$ is its own inverse modulo \$317.\$ (Your answer should be a list of integers greater than 0 and less than \$317,\$ separated by commas.)

Guest Sep 8, 2017
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To solve for the value of the residues A, we can use the formula:

A^2 = 1                 (mod prime)

This is only true for prime numbers. The given number which is 317 is a prime number therefore the values of the residues A are:

A = + 1, - 1

Since I believe the problem specifically states for the list of positive integers only and less than 317, a value of A = - 1 is therefore not valid. However, a value of – 1 in this case would simply be equal to:

317 – 1 = 316

Therefore the residue values A are 1 and 316.

cowgirlkatie03  Sep 8, 2017

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