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I have some proves But I want to prove them with deductive not inductive.
Here are my proves:

1) $2^{3n} - 1 $ is divisible by 7.

2) $2^n + (-1)^{n+1}$ is divisible by 3.

3) $n^2 + 2$ is not divisible by 4.

4) $11^n - 4^n$ is divisible by 7.

Is it possible to help me? (some of them is good too).
Thanks.

Amin
  • 595

1 Answers1

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  1. $2^{3n} - 1 = 8^n-1 = (8-1)a = 7a$ where a is the other factor.
    Hence, $2^{3n} - 1 $ is divisible by 7.
  2. $2^n + (-1)^{n+1} = 2^n - (-1)^n = [2-(-1)]b = 3b$ where b is the other factor.
    Hence,$2^n + (-1)^{n+1}$ is divisible by 3.
  3. $n^2 + 2 = 4k + 2 = 2(2k+1) = 2 \times \text{odd number}$, not divisible by $4$ ,when $n$ is even
    $n^2 + 2 = 8k+1+2 = 8k+3$, not divisible by $4$, when $n$ is odd
    Hence, $n^2 + 2$ is not divisible by 4.
  4. $11^n - 4^n = (11-4)c = 7c$ where c is the other factor.
    Hence, $11^n - 4^n$ is divisible by 7.
SchrodingersCat
  • 24,472
  • thanks. I can't vote for you because my reputation. just question about third one. when our number is even we use ${(2k)}^2 = 4k^2$ or we say $n^2$ is even so we use $n^2 = 2k$. How can we use $4k$ because how can we make number 2 with this format. (maybe k is from R not Z) Is it true? – Amin Oct 19 '15 at 18:42
  • Yes of course its true. $n^2=(2k)^2=4k^2=4m$ for $m=k^2$. And if you like my answer, you can accept it any time :) – SchrodingersCat Oct 19 '15 at 18:49
  • I accept that. thanks again. :) – Amin Oct 19 '15 at 18:56
  • You're welcome. – SchrodingersCat Oct 19 '15 at 18:56