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Actually I have done this problem by induction.(trivial)

Please tell me another method to do this problem instead of induction.

Souvik Dey
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user89071
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  • Modular arithmetic. Hands-on version: binomial expansion on $(5-2)^{2n+1}$. – anon Aug 06 '13 at 12:16
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    See "Why is $a^n−b^n$ divisible by $a−b$?": $\ $http://math.stackexchange.com/questions/188657/why-an-bn-is-divisible-by-a-b?lq=1 – Nikolaj-K Aug 06 '13 at 12:23

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Note: This solution assumes that the edit made by user Souvik Dey is in line with the OP's intent.

Note that $$ 3^{2n+1}=3\cdot 9^n\equiv 3\cdot(-1)^n\pmod{5} $$ and $$ 2^{2n+1}=2\cdot 4^n\equiv 2\cdot(-1)^n\pmod{5}. $$ So, we have $$ 3^{2n+1}+2^{2n+1}\equiv 5\cdot(-1)^n\equiv0\pmod{5}. $$

Nick Peterson
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HINT:

As for integer $a,b$ and integer $n\ge0,$ $a^n-b^n$ is divisible by $(a-b)$( Proof 1,2)

$a^{2n+1}+b^{2n+1}=a^{2n+1}-(-b)^{2n+1}$ is divisible by $a-(-b)=a+b$

Community
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lab bhattacharjee
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Let $A=3^{2n+1}+2^{2n+1}$.

  • $3=-2\pmod{5}$ hence $A=(-2)^{2n+1}+2^{2n+1}\pmod{5}$
  • $(-2)^{2n+1}=-2^{2n+1}$ hence $A=0\pmod{5}$

Likewise, $3^{2n}+2^{2n}=(-2)^{2n}+2^{2n}=2\cdot2^{2n}$ hence $5$ does not divide $3^k+2^k$ when $k$ is even, only when $k$ is odd.

Did
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By induction: For $n=0$ the result is clear.

Suppose the result true for $n$. $$3^{2(n+1)+1}+2^{2(n+1)+1}=9.3^{2n+1}+4.2^{2n+1}=4\left(3^{2n+1}+2^{2n+1}\right)+5.3^{2n+1}$$

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Since $$3\equiv-2\mod5,$$ we have $$3^{2n+1}\equiv(-2)^{2n+1}\equiv-2^{2n+1}\mod5,$$ and therefore $$2^{2n+1}+3^{2n+1}\equiv2^{2n+1}(1-1)\equiv0\mod5$$

Barry Cipra
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    Oh, I see Did posted the same solution while I was composing this one. We differ only in presentation (and Did's additional remark). – Barry Cipra Aug 06 '13 at 12:53