0

So, I have the following expression: $$(-5)\cdot{3}^m+(-3)\cdot{5}^m=(8k)\cdot3^m+(8k)\cdot5^m$$ ,where $k $ is an integer. I arrived at this by seeking for a contraddiction in a proof. Can I just equate the coefficients of the $3^m$ and $5^m$ terms and claim: $$8k=(-5) \;\;\;;\;\; 8k=(-3)$$ ? This is clearly impossible, otherwise also I know that $k$ must be negative so if I just move things around: $$-(3^m)(8k+5)=5^m(8k+3)$$ Then $(8k+5)$ and $(8k+3)$ would have opposite signs which once again, for integer $k$, it's trivially impossible. Are both these arguements valid? If they aren't, why?

dmtri
  • 3,270
Spasoje Durovic
  • 977
  • 1
    Not sure this is clear. If $m=1$ then the left hand is $-15-15=-30$ which is not divisible by $8$ – lulu Nov 09 '18 at 12:59
  • 2
    Your first argument is invalid $-$ you can't just equate coefficients like that. Your second argument, however, looks good to me. – TonyK Nov 09 '18 at 13:01
  • 1
    Please post the problem that this originates from. This will help us to determine if there are other mistakes before you got to this point . – Bill Dubuque Nov 09 '18 at 15:02
  • @BillDubuque That's not the point of my question, I don't want you to check the whole proof, just if that arguement it's fine, I was pretty sure the second arguement was valid, but I wanted to know if the first one was okay too, without much surprise, it isn't... If you're interessed in the problem: St. Petersburg olympiad from 1996 "Find all positive integers n such that" $$3^{n−1}+5^{n−1} \mid 3^{n}+5^{n}$$ After proving that n has to be odd I conjectured only n=1 was acceptable and after bashing through some algebra I arrived to what is posted up there... (n=1 is actually the only solution) – Spasoje Durovic Nov 09 '18 at 16:39
  • It is not unusual in questions like this that there is a prior error which prevents the method from applying. Moreover, knowing the original problem allows readers to suggest others methods - some of which may prove very illuminating. – Bill Dubuque Nov 09 '18 at 17:11
  • For OP, let $a_n = 3^n+5^n$. If $,a_{n-1}\mid a_n$ then $,a_{n-1}\mid a_n-3a_{n-1} = \color{#c00}2\cdot \color{#0a0}{5^{n-1}}.$ But $a_{n-1}$ is coprime to $\color{#0a0}{5^{n-1}}$ hence $,a_{n-1}\mid \color{#c00}2,$ thus $,\ldots$ – Bill Dubuque Nov 09 '18 at 17:16
  • See also here for a general formula for $\gcd(a^n+1,a^k+1)\ \ $ – Bill Dubuque Nov 09 '18 at 17:26

1 Answers1

1

This is too long for a comment. So I'll post it as an answer, though technically it is not.

Simply equating coefficients does not work here. For example $$3^0+5^0=2.3^0+0.5^0$$ Because both sets of the form $\{3^n\}_{n\in\mathbb{N}}$ and $\{5^n\}_{n\in\mathbb{N}}$ spans $\mathbb{N}=\{0,1,2,\cdots\}$ over $\mathbb{N}.$ It means you can write any integer as a finite linear combination of either elements of first set or the second set. These combinations (expansions) are unique if you impose the conditions that coefficients for first set are $\mathbb{Z}_3=\{0,1,2\}$ and coefficient for the second set are $\mathbb{Z}_5=\{0,1,2,3,4\}.$ For example $$123=3^4+3^3+3^2+2.3^1=4.5^2+4.5^1+3.5^0$$ These are just base $3$ and base $5$ expansions of $123 :)$

Bumblebee
  • 18,220
  • 5
  • 47
  • 87