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From K Rosen's Discrete Maths,

Theorem: If a and m are relatively prime integers and m > 1, then an inverse of a modulo m exists. Furthermore, this inverse is unique modulo m. (That is, there is a unique positive integer a less than m that is an inverse of a modulo m and every other inverse of a modulo m is congruent to a modulo m.)

Proof:
By Theorem 6 of Section 4.3,
because gcd(a, m) = 1, there are integers s and t such that
sa + tm = 1.
This implies that, sa + tm ≡ 1 (mod m).

Because t m ≡ 0 (mod m), it follows that sa ≡ 1 (mod m).
Consequently, s is an inverse of a modulo m.

I am struggling a lot and have gone through theorems given in the chapter multiple times, I'm unable to figure out how we can say,

sa + tm = 1.
This implies that, sa + tm ≡ 1 (mod m).

I realise that this a rookie question but if you can give me some link or even a theorem no. that would be of great help.

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    $\begingroup$ Cause $$sa+tm=1 \equiv 1 \pmod m$$ $\endgroup$ Commented May 8, 2016 at 6:58

2 Answers 2

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If $sa+tm=1$, then $sa+tm$ and $1$ are the same number. Thus, they will have the same remainder when divided by $m$, which is precisely what we mean when we say $sa + tm \equiv 1 \pmod{m}$.

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$sa + tm = 1 = 0 \cdot m + 1$

See? It is one greater than a multiple of $m$.

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