Prove that 33 divides 2^55 + 1

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The discussion revolves around proving that 2^55 + 1 is divisible by 33, as presented in a Mathematical Olympiad exercise. Participants explore various factoring techniques and polynomial expressions, noting that 33 can be expressed as 2^5 + 1. A suggestion is made to substitute x = 2^5 and rewrite the expression in terms of x to facilitate the proof. The conversation highlights the challenge of polynomial long division but ultimately leads to a successful resolution of the problem. The thread concludes with a prompt to generalize the findings to a^mn + 1 and explore conditions under which a^n + 1 is a factor.
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Homework Statement


The problem is exactly as stated in the title: "Prove that 255 + 1 is divisible by 33". This problem is an exercise from the Mathematical Oympiad Handbook in the section on factoring sums of powers.[/B]

Homework Equations


Results found so far in this section:
xn - 1 = (x-1)(xn-1+ xn-2 +...+x2 +x + 1)

x3 + 1 = (x+1)(x2- x + 1)

x3 + y3 = (x+y)(x2- xy +y2)

The Attempt at a Solution


So far, I have that 33 = 32 + 1 = 25 + 1 = (2+1)(24 - 23 + 22 -2 + 1) and that 255 + 1 = (2+1)(254-253+252+...-23 + 22 -2 + 1)

I do not know how to proceed from here. Presumably I'd want to show that the quotient of the two polynomials is an integer, but I don't know how to do that besides polynomial long division, and I doubt that's what the problem is trying to emphasize (and with so many terms, it would probably take forever, if it's even possible for polynomials of orders 5 and 55- not that I'd want to anyway, with so many terms).
 
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jack476 said:

Homework Statement


The problem is exactly as stated in the title: "Prove that 255 + 1 is divisible by 33". This problem is an exercise from the Mathematical Oympiad Handbook in the section on factoring sums of powers.[/B]

Homework Equations


Results found so far in this section:
xn - 1 = (x-1)(xn-1+ xn-2 +...+x2 +x + 1)

x3 + 1 = (x+1)(x2- x + 1)

x3 + y3 = (x+y)(x2- xy +y2)

The Attempt at a Solution


So far, I have that 33 = 32 + 1 = 25 + 1 = (2+1)(24 - 23 + 22 -2 + 1) and that 255 + 1 = (2+1)(254-253+252+...-23 + 22 -2 + 1)

I do not know how to proceed from here. Presumably I'd want to show that the quotient of the two polynomials is an integer, but I don't know how to do that besides polynomial long division, and I doubt that's what the problem is trying to emphasize (and with so many terms, it would probably take forever, if it's even possible for polynomials of orders 5 and 55- not that I'd want to anyway, with so many terms).
In that last expression, look at (254-253+252+...-23 + 22 -2 + 1).

Pair the first two terms 254-253 = 2(253) - 253 = ?

Similarly, pair the next two terms, etc.
Added in Edit:​

Actually there are some more straight forward ways to do this.

One of them is as follows:

Since you know that 33 = 25 + 1 , You can let x = 25 .

Write 255 + 1 in terms of x . See if you can show that x + 1 divides that result.
 
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SammyS said:
In that last expression, look at (254-253+252+...-23 + 22 -2 + 1).

Pair the first two terms 254-253 = 2(253) - 253 = ?

Similarly, pair the next two terms, etc.
Added in Edit:​

Actually there are some more straight forward ways to do this.

One of them is as follows:

Since you know that 33 = 25 + 1 , You can let x = 25 .

Write 255 + 1 in terms of x . See if you can show that x + 1 divides that result.

Brilliant, got it!

Thanks! =D
 
jack476 said:
Brilliant, got it!

Thanks! =D
Generalise to ##a^{mn}+1##. Under what circumstances will ##a^n+1## be a factor?
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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