Mathematical Induction and Algebra

Question

In a question taken from Discrete Mathematic With Applications

A question tries to prove $2^{2n}-1$ is divisible by 3.

In the solution it has $$2^{2k}(3+1)-1$$ $$2^{2k}.3+(2^{2k}-1)$$ What happened to the 1 in (3+1) ? are they saying 1 was equivalent to $2^{2k}$?

score 1 · Accepted Answer · answered Jul 27 '16 at 14:41

1

They just expanded the term with the first term by multiplying out the parentheses:

$$2^{2k}(3+1)-1=2^{2k}(3)+2^{2k}(1)-1=2^{2k}(3)+(2^{2k}-1)$$

answered Jul 27 '16 at 14:41

smcc

5,694

Wow .... my basic mathematics understanding are being tested in Induction . Thanks for answer – user2861799 Jul 27 '16 at 15:17

score 1 · Answer 2 · answered Jul 27 '16 at 15:05

1

$2^{2n}= 4^n$. But $4\equiv1\pmod3$ Hence $2^{2n}\equiv 1\pmod 3$.

answered Jul 27 '16 at 15:05

P Vanchinathan

19,504

Is it true that $4=1(mod 3)$ implies $4^n=1 ^n (mod 3)$? Take $n=2$. – Babai Jul 27 '16 at 15:30
A basic property of congruences is that $a\equiv b\pmod m,\ c\equiv d\pmod m$ implies $ac\equiv bd\pmod m$. – P Vanchinathan Jul 27 '16 at 15:41
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@ Babai: Yes $4^2=16$, when divided by 3 gives remainder 1. – P Vanchinathan Jul 27 '16 at 15:43
Oh yes, you are right! – Babai Jul 27 '16 at 15:45

score 0 · Answer 3 · edited Apr 13 '17 at 12:20

0

No, $1$ is not equivalent to $2^{2k}$. But the dsitributive law says that $2^{2k}(3+1)=3\cdot 2^{2k}+2^{2k}$. Now substract $-1$ from it. In case you want to prove that this is divisible by $3$ by induction, then see this MSE-question.

edited Apr 13 '17 at 12:20

Community

1

answered Jul 27 '16 at 14:41

Dietrich Burde

130,978

Mathematical Induction and Algebra

3 Answers3