Why $99x^2 \equiv 1 \mod 5 \implies (-1)x^2 \equiv 1 \mod 5$

Question

Lastly I have read a example of some exercise. There was this statement: $$99x^2 \equiv 1 \pmod 5\quad \implies\quad (-1)x^2 \equiv 1 \pmod 5$$ Can somebody explain that simple fact to me?

Apply the [Congruence Product Rule](https://math.stackexchange.com/a/879262/242), using $\ 99\equiv 9\equiv -1\pmod{5}\ \ $ — Bill Dubuque, May 10 '19 at 15:58

score 2 · Answer 1 · answered May 10 '19 at 15:46

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Since we have $$99\equiv -1 \mod 5$$ this is the reason.

answered May 10 '19 at 15:46

Dr. Sonnhard Graubner

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score 2 · Answer 2 · answered May 10 '19 at 15:46

2

Because $99\equiv 4 \equiv -1 \mod 5$

answered May 10 '19 at 15:46

pwerth

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score 2 · Answer 3 · answered May 10 '19 at 15:47

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$$99\equiv 4=(5-1)\equiv -1 \pmod{5}$$

:)

answered May 10 '19 at 15:47

Kandinskij

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score 2 · Accepted Answer · answered May 10 '19 at 15:49

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$$99x^2\equiv_5100x^2-x^2\equiv_50-x^2\equiv_5-x^2$$

answered May 10 '19 at 15:49

cansomeonehelpmeout

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Why $99x^2 \equiv 1 \mod 5 \implies (-1)x^2 \equiv 1 \mod 5$

4 Answers4