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This question is for all $a,m$ and $n$ $\in \mathbb{N}$

I'm really stuck on this question, I tried splitting the modulus and doing the congruence one at a time but can't find a way to put it back using CRT.

Because

$(a+1)^n \equiv 0\pmod {a+1}$ and $a^m \equiv 0\pmod a$

Also, I found out by trying different integers that

$(a+1)^n \equiv 1\pmod a$

Will this be useful for proving the question?

blablabla
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1 Answers1

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Hint $\rm\ LHS-RHS\equiv 0\,$ mod $a$ and mod $a\!+\!1$ so it is divisible by $\,a\,$ and $\,a\!+\!1\,$ thus also by their lcm = product = $\,a(a\!+\!1)\,$ since they are coprime. Note that mod $\ a\!+\!1\!:\ a\!+\!1\equiv 0\,\Rightarrow\, a\equiv -1.$

Or, using CRT verify that both sides are solutions of $\, x\equiv 1\pmod{a},\ x\equiv (-1)^m\pmod{a\!+\!1}\,$ hence, by CRT uniqueness, they are congruent $\!\pmod{a(a+1)}\,$

Quite frequently, uniqueness theorems provide powerful tools for proving equalities.

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Bill Dubuque
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  • For the second solution, how did you get the solutions $ x\equiv 1\pmod{a},\ x\equiv (-1)^m\pmod{a!+!1}\ $ ? And how does it lead to congruence by CRT? I thought you can only merge the (mod a(a+1)) if both solutions are common. – blablabla Nov 15 '16 at 04:22
  • @jasonderulo $\ x_1 := a^m + (a+1)^n\equiv (-1)^m + 0^n\pmod{a+1}.,$ Similarly $,x_1\equiv 1\pmod a.,$ Thus $,x_1,$ is a solution of said system of congruences. Similarly, $,x_2 := 1+a(1+(-1)^{m-1}),$ is a solution of the system. Thus $,x_1 \equiv x_2\pmod{a(a+1)},$ by CRT uniqueness. – Bill Dubuque Nov 15 '16 at 04:39