Show $2^{55} + 1$ divisible by 33 using only basic identities

Question

I'm working my way through The Mathematical Olympiad Handbook by A. Gardiner.

In the section A little useful mathematics right at the start the following basic identities are laid out

$x^n -1 = (x-1)(x^{n-1} + ... + 1)$
$x^n - y^n = (x-y)(x^{n-1} + ... + y^{n-1})$
$x^n + 1 = (x+1)(x^{n-1} + ... + 1)$
$x^n + y^n = (x-y)(x^{n-1} + ... + y^{n-1})$

In this context the following two exercises are posed

Prove $2^{55} + 1$ is divisible by $33$
Prove $1900^{1990} -1$ is divisible by $1991$

So for (1) the strong implication seems to be to somehow use the fact that $33=2^5+1=(2+1)(2^4-2^3+2^2-2^1+1)$ and use that, but other than realising that I am slightly lost.

I can see using Fermat's Little Theorem allows an attack ($2^{55}=(2^5)^{11}$ etc.) but how to do it with just those identities I can't see it as Fermat's Little Theorem is used later as a second proof of this problem in the number theory section.

Can anyone see how to do it?

@Peter I believe OP wants it done via the "basic identities that are laid out" rather than "Fermat's Little Theorem that comes later". — Calvin Lin, Dec 19 '20 at 14:30
@CalvinLin The approach also works without FLT. I admit that here we can do better. — Peter, Dec 19 '20 at 14:31
@Peter Interesting. I get the $ 2+1 = 3$. Can you write up how to show it is divisible by 11 (without showing 33 directly)? — Calvin Lin, Dec 19 '20 at 14:32
@Peter My point is that you're factoring out $2^5 + 1 = 33$ (IE using the basic identities), so the hint of "show divisibility by 3 and by 11" suggests doing them completely independently of each other, vs the approach suggested by the solutions below. $\quad$ Apart from that, we're saying the same thing. — Calvin Lin, Dec 19 '20 at 14:36
Hint: in $,X-Y\mid X^n - Y^n$ specialize $,X,Y = x^k,y^k,$ to the needed powers. Same for $,X^n + Y^n.,$ More generally we can use the Polynomial Factor Theorem. It is easier using equations (congruences) vs. relations (divisibility), see the Polynomial Congruence Rule. — Bill Dubuque, Dec 19 '20 at 19:14

score 5 · Answer 1 · answered Dec 19 '20 at 14:28

5

Hint:

$$2^{55} +1 =(2^5)^{11} +1 = 32^{11} +1 = (32+1)(\dots) $$ using your third identity.

answered Dec 19 '20 at 14:28

Vishu

14,469

Thanks seems so obvious now. I am struggling to use a similar technique to crack $1900^{1990}-1$ divisible by 1991. Any idea? – Gau55man Dec 19 '20 at 14:40
1

@Gau55man Have you copied down the problem correctly? Wolfram says that is not true https://www.wolframalpha.com/input/?i=+Is+%281900%5E%281990%29-1%29%2F1991+an+integer – Vishu Dec 19 '20 at 14:53
Yes I've triple checked $1900^{1990} -1$ is divisible by 1991. Most sensible pattern would mean what was intended was $1990^{1990}-1$ is divisible by $1991$ I'm guessing. – Gau55man Dec 19 '20 at 16:38
@Gau55man What have you checked it with? – Vishu Dec 19 '20 at 18:01
I meant I tripled checked that is what the questions says so there is a typo/mistake in my book. I'm assuming intention is $1990^{1990} -1$ which admits solution by recognising as others have that $(1990^2)^{955} -1 = (1990^2 - 1)(...)$ then using identity $n^2 -1=(n+1)(n-1)$ which gives $1990^2 -1=1991 \cdot 1989$ showing that 1991 is a factor. – Gau55man Dec 19 '20 at 22:35
@Gau55man Yes, that is probably the case. – Vishu Dec 20 '20 at 09:54

score 3 · Answer 2 · answered Dec 19 '20 at 14:28

3

$$(2^{5})^{11} + 1 = 32^{11} + 1 \overset * \equiv (-1)^{11} + 1 = 0 \pmod{33}$$ $(*)$: $32 = 33 - 1 \equiv -1 \pmod {33}$

answered Dec 19 '20 at 14:28

VIVID

11,604

user376343 · Answer 3 · 2020-12-19T14:58:28.587

1

For divisibility of $2^{55}+1$ by $33,$ we can write $$2^{55}+1=(2^{5})^{11}+1=(33-1)^{11}+1,$$ and apply the third identity.

Isn't there a typo in the second exercise? I suggest to prove divisibility of $1990^{1990} -1$ by $1991$ instead. We have
$$1990^2 -1=1991\times 1989$$ and use the first identity.

edited Dec 19 '20 at 14:58

answered Dec 19 '20 at 14:43

user376343

8,311

1

$1900^2-1 \ne 1991\times 1989 $ – Vishu Dec 19 '20 at 14:53
1

You're right. I was rewriting it while you were posting your comment. Then I noticed you also think there is a typo in OP. – user376343 Dec 19 '20 at 15:01

Show $2^{55} + 1$ divisible by 33 using only basic identities

3 Answers3