n^2|a^n-b^n

rezashiri · 1389/8/11

اگر داشته باشیم

$a\equiv b(mod(n))$

ثابت کنید :

$a^{n}\equiv b^{n}(mod(n^{2}))$

AidinT · 1389/8/11

$a^n - b ^ n = (a-b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + ... + b^{n-1})$

$a\overset{n}\equiv b \rightarrow a^{n-1-i}b^i \overset{n}\equiv a^{n-1-i}a^i \overset{n}\equiv a^{n-1}$

$\rightarrow (a^{n-1} + a^{n-2}b + a^{n-3}b^2 + ... + b^{n-1}) \overset {n}\equiv n(a^{n-1})$

$\rightarrow (a^{n-1} + a^{n-2}b + a^{n-3}b^2 + ... + b^{n-1})(a-b) \overset{n^2}\equiv nq.nq' \overset{n^2}\equiv 0\rightarrow a^n \overset{n^2}\equiv b^n$

mrbayat · 1389/8/11

$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots +b^{n-1})$

$\begin{Bmatrix}a-b\overset{n}{\equiv }0\\ a^{n-1}+a^{n-2}b+\cdots +b^{n-1}\overset{n}{\equiv }na^{n-1}\overset{n}{\equiv }0 \end{Bmatrix}\Rightarrow a^n-b^n\overset{n^2}{\equiv }0$

n^2|a^n-b^n

rezashiri

Well-Known Member

AidinT

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mrbayat

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