SOLUTION: Let $a$ be an integer such that $0 \le a \le 10$ and $a^2 \equiv a \pmod{11}$. If $a \neq 0,$ then find the value of $a$.

Question 1207670: Let $a$ be an integer such that $0 \le a \le 10$ and $a^2 \equiv a \pmod{11}$. If $a \neq 0,$ then find the value of $a$.
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52776) (Show Source): You can put this solution on YOUR website!
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The answer is a = 1.

This answer is OBVIOUS.

Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!

Since we can effectively "divide" both sides of by 'a' to arrive at .
Technically it's not division that's happening, but instead multiplication of the multiplicative inverse.

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Another pathway is
(mod 11)

(mod 11)

(mod 11)

That leads to either a = 0 or a-1 = 0
But is specified in the instructions, so we must conclude that a-1 = 0 leads to a = 1.

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Yet another alternative method is to simply brute force your way through things.
Try a = 1, a = 2, etc all the way up to a = 10.

a = 1
a^2 = a (mod 11)
1^1 = 1 (mod 11)
1 = 1 (mod 11)
That works out

a = 2
a^2 = a (mod 11)
2^2 = 2 (mod 11)
4 = 2 (mod 11)
That doesn't work out since the two sides don't boil down to same number, and we cannot reduce either side.

a = 3
a^2 = a (mod 11)
3^2 = 3 (mod 11)
9 = 3 (mod 11)
That doesn't work out either.

a = 4
a^2 = a (mod 11)
4^2 = 4 (mod 11)
16 = 4 (mod 11)
5 = 4 (mod 11)
This is the first time that we reduce mod 11 on the left hand side.
You can repeatedly subtract 11 from the item until getting in the range of between 0 and 10, or you can divide by 11 to look at the remainder.
Like the previous cases, the two sides don't match, so a = 4 isn't a solution.

Keep this process going and you'll find that the values a = 5 through a = 10 are also non-solutions.

This confirms that only a = 1 is the solution.

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