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Let $(a,b)$ denote the greatest common divisor of $a$ and $b$.

With $ \ d = (51^{\large 610}\! + 1,\, 51^{\large 671}\! −1)$

and $\ \ x \,=\, (d^{\large 671} + 1,\, \ d^{\large 610} −1 )$

find $\ X = (x\bmod 10)$

I used $y=51^{61}$ to reduce $d$ to $d=(y^{10}+1,y^{11}-1) = (y^{10}+1,y+1)$.

What should I do now?

Bill Dubuque
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rj123
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  • Were you previously taught any method to find gcds of the form $,(a^j+1,a^k-1)$? Does "trick" mean a method of computing said gcd $!\bmod 10$ that is (much) simpler than computing the gcd then reducing it $\bmod 10$? – Bill Dubuque Mar 19 '19 at 15:54
  • As a further hint, note that if $\ y=x^{61}\ $, then $\ x^{610}+1= y^{10}+1\ $ and $\ x^{671}-1=y^{11}-1\ $. – lonza leggiera Mar 19 '19 at 16:09
  • The reduction to the case of coprime exponents in the prior comment of @Ionza works generally, e.g. see this answer. We can handle this problem in a very similar manner as I do there. See also this answer. – Bill Dubuque Mar 20 '19 at 00:27
  • Yeah i want to know the method for computing $(a^{j}+1,a^{k}-1)$ I mean how to derive this formula @BillDubuque – rj123 Mar 21 '19 at 09:55
  • And I do need to find d at first @BillDubuque – rj123 Mar 21 '19 at 10:06
  • You can use the method in the answers I linked. – Bill Dubuque Mar 21 '19 at 12:32
  • I have reduced the given d to $y=x^{61}$ then d reduced to this $d=(y+1,y^{10}+1) $ what should I do know? @BillDubuque – rj123 Mar 21 '19 at 12:35
  • @q123 Follow the linked answers. Recall $,(a,b) = (a,,b\bmod a),$ so $,(y+1,f(y)) = (y+1,f(-1)),$ by $\bmod y+1!:,\ y\equiv -1,\Rightarrow, f(y)\equiv f(-1),,$ for any polynomial $f(y)$ with integer coeff's. – Bill Dubuque Mar 22 '19 at 20:03
  • I posted an answer with a full proof (which uses a common theorem $\bf T1$ for both gcds). – Bill Dubuque Mar 22 '19 at 23:41
  • Were any of the answers helpful? If you need help understanding them then you can pose questions in comments on the answers. – Bill Dubuque Mar 25 '19 at 15:51

3 Answers3

1

First note: $\gcd(a^m \pm 1, a+1)=\gcd((a^{m}\pm 1)-(a^{m}+a^{m-1}),a+1) = \gcd(a^{m-1}\mp 1, a+1)$

And via induction $\gcd(a^{m}+1, a+1) = \gcd(2, a+1)$ if $m$ is even. $\gcd(a^{m} - 1,a+1) =\gcd(0, a+1) = a+1$ if $m$ is even. (And the opposite results if $m$ is odd).

so if we let $a= 51^{61}$.

Then $\gcd(51^{610} + 1, 51^{671} - 1)=$

$\gcd(a^{11}-1,a^{10} + 1)=$

$\gcd((a^{11} -1)-(a^{11} + a), a^{10} + 1) =$

$\gcd(a^{10} + 1, a+ 1) = 2$

...

Let $b = 2^{61}$ and so

$\gcd(2^{671}+1, 2^{610} -1)= \gcd (b^{11} + 1, b^{10} -1)=$

$\gcd((b^{11}+1)-(b^{11}-b), b^{10}-1)=\gcd(b^{10}-1,b+1)=b+1= 2^{61}+1$

fleablood
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As in the proofs here and here, we reduce to coprime powers then apply the $\rm\color{#90f}{Euclidean}$ algorithm.

$a = 51^{\large 61}\Rightarrow\, d = (a^{\large 11}-1,\,a^{\large 10}+1) = (a\!+\!1,2) = 2\,$ by $\,\bf T1\,$ below, with $\ s = -1$

$a\, =\, d^{\large 61}\Rightarrow\,x = (a^{\large 11}+1,\,a^{\large 10}-1) =\, a\!+\!1 = d^{\large 61}\!+1 = 2^{\large 61}\!+1\,$ by $\,\bf T1,\,$ $\,s = 1$

${\bf T1}\,\ (s,a)\! =\!1\, \Rightarrow\, (a^{\large 11}\!+s,\,a^{\large 10}-s)\, = (a\!+\!1,\,1\!-\!s).\ $ Proof: $\,\rm\color{#90f}{using}$ $\ (x,y) = (x,\, y\bmod x)$

$\begin{align} (\color{#0a0}{a^{\large 11}}\!+s,\,{a^{\large 10}}\!-s) &= (\color{#0a0}{s}(\color{#0a0} a\!+\!1),\, {a^{\large 10}}\!-s)\ \ \ \,{\rm by}\ \ \bmod a^{\large 10}\!-s\!:\,\ a^{\large 10}\!\equiv s\,\Rightarrow\, \color{#0a0}{a^{\large 11}}\!\equiv a^{\large 10}a \equiv \color{#0a0}{sa} \\[.2em] &= \ \ \ \, (a\!+\!1,\ \,\color{#c00}{a^{\large 10}}\!-s)\ \ \ \ {\rm by}\,\ \ (s,\,a^{\large 10}\!-s) = (s,a^{\large 10})=1, \ \, {\rm by}\,\ \ (s,a) = 1\\[.2em] &=\ \ \ \ (a\!+\!1,\ \ \ \color{#c00}1\, -\, s) \ \ \ \ {\rm by}\ \ \bmod a+1\!:\ \ \ \ a\equiv -1\,\Rightarrow\, \color{#c00}{a^{10}}\equiv (-1)^{10}\equiv\color{#c00} 1 \\[.2em] \end{align}$

Bill Dubuque
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  • Thus $,X = (2^{\large 61}!+1\bmod 10) = 3,\ $ by $\ 2^{\large 61}! \bmod 10 = 2\big((2^{\large 4})^{\large 15} \bmod 5\big) = 2\ \ $ – Bill Dubuque Mar 22 '19 at 23:29
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$$51^{671}=51^{610}\times 51^{61}-1=(51^{610}+1)51^{61}-(51^{61}+1)$$

$$(51^{61}, 51^{61}+1)=1$$

So we may write:

$$d=(51^{610}+1, 51^{671}-1)=(51^{610}+1, 51^{61}+1)$$

$$51^{61}+1=52 k=2\times 26 k$$

$$51^{610}+1=52 k_1 +2=2(26 k_1+1)$$

$$(26, 26k_1+1)=1$$

However $k$ and $26 k_1+1$ may have common divisors. If we assume $d=2$ then we have:

$$x=(2^{671}+1, 2^{610}-1) $$

$$2^{671}+1=(2^{610}-1)2^{61} +2^{61}+1$$

$(2^{61},2^{61}+1)=1$, Therefore:

$$x= 2^{61}+1$$

sirous
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