How to calculate the $\gcd(10^6 +1 , 10^2 +1)$?

Question

I can't really figure out how to approach this question..

I have tried to factor 10^2 +1 out of 10^6 +1 , however, the '+1' part makes it difficult.

$10^6 + 1 = (10^2 +1) \cdot 10^3 + 899000$,
$10^2 + 1 = 899000 \cdot ...? $

Here is where I get stuck, because $899000$ is way bigger than $10^2 + 1$..

Does anyone know how to approach this?

Thanks!

https://math.stackexchange.com/questions/641443/proof-of-anbn-divisible-by-ab-when-n-is-odd — lab bhattacharjee, Nov 06 '17 at 13:54

score 6 · Accepted Answer · answered Nov 06 '17 at 13:54

6

Use $$x^3+1=(x+1)(x^2-x+1).$$ I think it must help.

answered Nov 06 '17 at 13:54

Michael Rozenberg

194,933

How it can help? – jonsno Nov 06 '17 at 13:56
@samjoe It can help if you plug in the appropriate value of $x$, such that two of the expressions become $10^2+1$ and $10^6+1$... – 5xum Nov 06 '17 at 13:57
But how does it ensure that $(x+1)$ is indeed the greatest common divisor? – Gaurang Tandon Nov 06 '17 at 13:57
@samjoe We get already that $10^6+1$ is divided by $10^2+1$. – Michael Rozenberg Nov 06 '17 at 13:57
@Gaurang Tandon What is a greater number than $n=10^2+1$ for which $10^2+1$ is divisible by $n$? – Michael Rozenberg Nov 06 '17 at 14:00
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@MichaelRozenberg Of course, none :) My query is satisfied. Thanks! – Gaurang Tandon Nov 06 '17 at 14:01

score 5 · Answer 2 · answered Nov 06 '17 at 13:54

5

Since $10^2+1=101$ is a prime number, you only have two options:

$101$ divides $10^6+1$, in which case the gcd is $101$
$101$ does not divide $10^6+1$, in which case the gcd is $1$.

answered Nov 06 '17 at 13:54

5xum

123,496
6
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Thanks for your answer!
How can I find which one of the two options it is?
Or is this just trying to factor 101 out of 10^6+1 – Math420 Nov 06 '17 at 14:13
@R3E1W4 Dividing $1000001$ by $101$ isn't all that hard, is it? – 5xum Nov 06 '17 at 14:14
Ah I see... haha thanks! – Math420 Nov 06 '17 at 14:15
@R3E1W4 I really think you should accept the other answer (by Michael Rozenberg). It's much more elegant than mine. – 5xum Nov 06 '17 at 14:40

score 0 · Answer 3 · answered Nov 14 '17 at 06:08

Well, the "throw rocks at it until it falls off a cliff approach":

$\gcd (10^6 + 1, 10^2 + 1) = \gcd([10^6 + 1] - 10^4(10^2 + 1), 10^2 + 1)$

$=\gcd (-10^4 + 1, 10^2 + 1) = \gcd(-10^4 + 1 + 10^2(10^2 + 1), 10^2 + 1)$

$= \gcd(10^2 + 1,10^2+1) = 10^2 + 1$.

....

There's also the insightful. $x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + .... + x + 1)$ and if $n$ is odd then $x^n + 1 = (x+1)(x^{n-1} - x^{n-2} + .... + x^2 - x + 1)$ and let $x = 10^2$ and $n = 3$ and... oh,... that's Mike Rozenberg's elegant answer... isn't it?

How to calculate the $\gcd(10^6 +1 , 10^2 +1)$?

3 Answers3