I have found out, that the following is true for modular arithmetic when $t$ is a natural number.
$$a^t \bmod\ n \equiv (a\bmod\ n)^t\bmod\ n$$
But I have been unable to find a proof for this, does anyone have a source that proves this conjecture?
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Hint $$(a+cn)(b+dn)-ab=n(ad+bc+cdn).$$ – awllower Feb 02 '14 at 16:08
2 Answers
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Hint
$$a^t-b^t=(a-b)(a^{t-1}+a^{t-2}b+\cdots+ab^{t-2}+b^{t-1})$$
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I'm afraid I don't understand. $$(a^{t−1}+a^{t−2}b+⋯+ab^{t−2}+b^{t−1})$$what pattern does it follow, ie. what is the "..."? – Clewett83 Feb 02 '14 at 16:19
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From the given equality if $n$ divides $a-b$ it divides also $a^t-b^t$. Right? – Feb 02 '14 at 16:23
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But, how is that relevent for what we are trying to prove, there is no substraction? – Clewett83 Feb 02 '14 at 16:42
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We want to prove $$a^t \bmod\ n \equiv (a\bmod\ n)^t\bmod\ n.$$ Let $ a \bmod n \equiv b $.
Thus by substitution we are proving $$a^t \bmod\ n \equiv b ^t\bmod\ n.$$ But this means we need to show $ n \mid a^t - b^t $ under the assumption that $ n \mid a - b $ .
Then using the fact that $a^t-b^t=(a-b)(a^{t-1}+a^{t-2}b+\cdots+ab^{t-2}+b^{t-1})$
john
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