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This one should be quite easy, I tried with mathematical induction but the things started to complicate so I would like to see how someone of you there will prove this:

Prove that $$a_k=(18k+1)2^{18k+1}+1$$ is composite number for every $k\gt 0$.

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Note that $a_0$ is not composite. Is the condition supposed to be $k \gt 0$? –  hardmath Jan 30 '13 at 9:23
    
Yes. Good observation. –  A.P. Jan 30 '13 at 10:00

2 Answers 2

up vote 7 down vote accepted

Go $\pmod 3$ and see what happens. However your mouse over the gray area below for the complete solution.

$$2 \equiv -1 \pmod3 \implies 2^{18k} \equiv 1 \pmod3 \implies 2^{18k+1} \equiv -1 \pmod3$$Also, $$18k+1 \equiv 1 \pmod 3$$Hence,$$(18k+1)2^{18k+1} \equiv -1 \pmod3 \implies (18k+1)2^{18k+1} + 1 \equiv 0 \pmod3$$

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What lead you to work (mod 3), as opposed to any other choice? –  User58220 Jan 31 '13 at 5:12
    
@user58220 To prove a sequence of numbers are composite, probably the most simplest way is to prove that all of them are divided by a fixed number. The sequence here is clearly not divisible by $2$. So the next bet is $3$. –  user17762 Jan 31 '13 at 5:25

$\rm (a\!-\!1,b\!+\!1)\,|\,\color{#C00}{a}\,\color{#0A0}{b}^{2n+1}\!+1\:$ since $\rm\,mod\ (a\!-\!1,b\!+\!1)\!:\: a\equiv 1,\, b\equiv -1\:$ so it's $\rm\,\equiv\, \color{#C00}{1}\,\color{#0A0}{(-1)}^{2n+1}\!+1\equiv\, 0$

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Welcome back Bill! –  user17762 Jan 30 '13 at 17:55

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