Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $k\gt 0$.

share|cite|improve this question
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
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,$$(18k+1)2^{18k+1} \equiv -1 \pmod3 \implies (18k+1)2^{18k+1} + 1 \equiv 0 \pmod3$$

share|cite|improve this answer
What lead you to work (mod 3), as opposed to any other choice? – DJohnM 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 $\rm\,\equiv\, \color{#C00}{1}\,\color{#0A0}{(-1)}^{2n+1}\!+1\equiv\, 0$

share|cite|improve this answer
Welcome back Bill! – user17762 Jan 30 '13 at 17:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.