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...a((b1 + b2 + ... + bn) + b(n+1)) = = a(b1 + b2 + ... + bn) + ab(n+1) = ab1 + ab2 + ... +

**abn**+ ab(n + 1), where the second equality comes from associativity and the basis...2 Answers · Science & Mathematics · 22/03/2010

...there are integers m and n such that 1=xm+an. Then b=xbm+

**abn**( multiply both sides by b) b=xbm+(xk)n (since ab=xk) b=x(bm+kn) ...2 Answers · Science & Mathematics · 28/03/2010

... - bⁿ) = (a - b) (aⁿ⁻¹ + aⁿ⁻²b + . . . +

**abⁿ**⁻² + bⁿ⁻¹) So we can also factor x⁴ - 1...4 Answers · Science & Mathematics · 17/09/2010

n = 4 a^4-b^4 = (a^2+b^2)(a+b)(a-b)

2 Answers · Science & Mathematics · 14/02/2010

go to mathnerds.com and you will get your answer.

1 Answers · Science & Mathematics · 16/01/2010

I think you have to use the algebraic structure of G/N as well. Your argument shows that 1, a, a^1, ..., a^m can not represent pairwise distinct cosets, but it doesn't show that a^m is in N. (For instance, your argument doesn't exclude that 1, a, a^2, ..., a^m...

2 Answers · Science & Mathematics · 24/09/2011

Lets assume that no two sides a b and c are equal, and a > b > c > 0 Then a/b > 1 and b/c > 1 and there exists 0.001 > ε > 0 such that a/b > 1+ ε and b/c > 1 + ε Lets chose n ~ 1/ε, then (1+ε)ⁿ ~ e = exp(1), and aⁿ...

4 Answers · Science & Mathematics · 05/10/2008

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