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From a book I knew something about RSA algorithm .There I found a modular arithmetic property i.e $ (a\bmod n)^d\mod n=a^d\bmod n$

I don't know why this property works .Can anyone give me an intuitive proof for above property?

Michael Hardy
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Shahed al mamun
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  • Learn about congruences. All will become clear (and afterwards you never again need to write more than one mod per line). – Jyrki Lahtonen Jun 04 '16 at 18:46
  • Exponentiation (to a fixed power) is the same as repeated multiplication. Do you know why ($a$ mod $n$)($b$ mod $n$) mod $n$ = $ab$ mod $n$? – Erick Wong Jun 04 '16 at 18:47
  • See for example the calculation here. Learn congruences and join us! You will never want to go back to using mod as meaning the remainder of integer division. It is a comparison operator! – Jyrki Lahtonen Jun 04 '16 at 18:58

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Let's pretend $a'=a+k n$ for some integer $k$, hence $a' \equiv a \pmod n$

We have $$(a')^d=(a+kn)^d=\sum_{i=0}^{d} \binom{d}{i}a^i(kn)^{d-i}\equiv a^d \pmod n$$

Aritra Das
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Evariste
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