0

I implemented Extended Euclid Algorithm in c++ to solve this problem. Any approaches that you could it by hand.

$\gcd(61^{610}+1,61^{671}-1)=\ ?$

Thanks in advance.

YuiTo Cheng
  • 4,705
80567
  • 21
  • 5
  • 6
    "I used Extended Euclid Algorithm to solve this problem" -- ... is this not a mathematical approach?? – PrincessEev Mar 21 '19 at 08:46
  • 2
    Do some research on Chinese Remainder Theorem. Moreover, as a hint: $ 610+61=671$. – Stan Tendijck Mar 21 '19 at 08:55
  • @EeveeTrainer I programmed it in c++, – 80567 Mar 21 '19 at 09:28
  • 1
    ... and? That doesn't mean it's not inherently mathematical. If you were determined enough, you could've done it by hand - computers are just faster and less prone to mistakes/distraction/bodily needs. You're going to have to make clearer what you mean by this "some mathematical approach" thing if you want any meaningful help. – PrincessEev Mar 21 '19 at 09:29
  • Sub-problem of a question a day prior (with $51$ changed to $61$, which makes no difference). – Bill Dubuque Mar 22 '19 at 20:07

1 Answers1

3

You should definitely use Euclid's algorithm (but you won't need the extended version); it's just you also need some more insight into the numbers you're working with. Define $n:=61^{61}$ so your problem is computing $$(n^{11}-1,\,n^{10}+1)=(n^{10}+1,\,n+1)=(n+1,\,2)=2.$$

J.G.
  • 115,835
  • $$61^{610} , 61^{671}$$ are odd so $$61^{610} + 1, 61^{671} - 1$$ are even and we can say that 2 is a factor of both, but to prove 2 is gcd I have to prove that there is no other common factor how should i prove it? – 80567 Mar 21 '19 at 09:47
  • @ChandanKumar I just showed you how. Since $d$ is a common divisor of $ka+r,,a$ iff it is of $a,,r$, these pairs have the same gcd. Therefore, we eventually reduce the problem to $(n+1,,2)$, which is $2$ because $2$ isn't big enough to have any larger factors. – J.G. Mar 21 '19 at 09:55
  • 1
    thanks a lot for this elegant solution. – 80567 Mar 21 '19 at 10:00
  • @Chandan You can do such problems very easily if you master modular arithmetic, e.g. see this answer and the answers linked there. – Bill Dubuque Mar 23 '19 at 01:27