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I have a question from a pattern that I noticed in numbers of the form: $$I_n=2^{2+4n}+1$$ for $n\geq0$.

All $I_n$ numbers up to $n=7$ seem to show $5\mid I_n$. But is this the case for any arbitrary $n$?

I tried to show this, but didn't get too far. Some information I found useful was:

  • $I_n$ is clearly an odd number, so for $5\mid I_n$ to hold, its last digit must be $5$.
  • $\{I_n\mid n\in\mathbb{N}\}=\{4^e+1\mid e\in\Bbb N\text{ is odd}\}$.
HeroZhang001
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Simón Flavio Ibañez
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3 Answers3

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Observing the first few items of the sequence $\{2^n\}$, $$2,4,8,16,32,64,128,256,512,1024,\cdots,$$ you will find that the single digits of these items are $$2,\boxed 4,8,6,2,\boxed 4,8,6,2,\boxed 4,\cdots,$$ respectively. Therefore by induction, the desired result clearly holds.

HeroZhang001
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    That is not a valid proof by induction. There are plenty of patterns that hold for small initials segments but later fail, e.g. see here. – Bill Dubuque May 16 '24 at 16:41
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque May 16 '24 at 16:43
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    @BillDubuque Obviously the single digit of the product of two integers must be the single digit of the product of the single digits of these two integers, so the proof is valid. – HeroZhang001 May 16 '24 at 16:51
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    @BillDubuque To see this, let $a=10^na_n+\cdots+10a_1+a_0$, $b=10^nb_n+\cdots+10b_1+b_0$, where $a_1,\cdots,a_n,b_1,\cdots,b_n=0,1,\cdots,9$, then $ab=(10^na_n+\cdots+10a_1+a_0)(10^nb_n+\cdots+10b_1+b_0)$. The single digit of $ab$ is the single digit of $a_0b_0$. – HeroZhang001 May 16 '24 at 16:57
  • Any justification of your claim should be placed in the answer, not in a comment. – Bill Dubuque May 16 '24 at 17:16
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For every natural number $n$:

$$a - b \mid a^n-b^n$$

For every odd natural number $m$: $$a + b \mid a^m + b^m$$

Now, let's say: $a=4$ and $b=1$, then we get:

$$4 + 1 \mid 4^m+1^m$$

We all know that $1^m=1$, hence:

$$4 + 1 \mid 4^m+1$$

The general way of writing an odd number $m$ is: $m=2n+1$ (where $n$ is a natural number, of course), hence:

$$4 + 1 \mid 4^{2n+1}+1$$

Now, replace $4 + 1$ by $5$ and $4$ by $2^2$ or $4^{2n+1}$ by $2^{4n+2}$ and that's it :-)

Dominique
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Are you familiar with proof by Induction? Prove something is true for a starting case. Prove if it is true for one integer, it is true for the next.

$I_n=2^{2+4n}+1$

$I_0=5$

$I_{n+1}=2^{6+4n}+1=16I_n-15$

So $5|I_n\implies 5|I_{n+1}$

TurlocTheRed
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  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque May 16 '24 at 15:51
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    Sorry. I think I'm not sure what constitutes a duplicate. I agree those linked solutions will work. I got the impression OP is unfamiliar with congruences and variable bases given his focus on numbers ending in 5. Is prob in early weeks of a discrete math class. That suggests 3 of the 4 linked answers wouldn't help. The base 7 Question's solutions should be adequate, to demonstrate using induction, supposing OP knows there is work that precedes the induction step which I'm on the fence about taking for granted. – TurlocTheRed May 16 '24 at 16:28
  • The key arithmetical idea in the inductive step is much clearer when the proof is reorganized as below

    $$\bmod 5!:\ \color{#c00}{4^2\equiv 1},,f_n \equiv 4^{1+2n}\Rightarrow, f_{n+1} \equiv \color{#c00}{4^2} f_n \equiv f_n\qquad\qquad$$ Thus $,f_n\equiv f_0\equiv 4,$ so $,f_n+1\equiv 0.,$ The dupes (and their links) have essentially every common way to prove this. $\ \ $

     – Bill Dubuque May 16 '24 at 18:10