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I would like to know if there is a mathematical approach to finding the LCM of $(29^{17} +2 , 29^{17} -1)$?

Even if we would rearrange it to a fraction of the form $\frac{(29^{17} +2)\cdot (29^{17} -1)}{gcd(29^{17} +2 , 29^{17} -1)}$ , we would still need to calculate the GCD. Is there a way using number theory that I am missing? I dont want to resort to using calculator to figure this one out.

If its not possible to find the LCM, is it possible to find just it's unit digit?

Ray Penber
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    Have you tried the Euclidean algorithm to find the $\gcd$? – Arthur Oct 24 '19 at 08:06
  • @Arthur no i haven't tried it. Since this question has been posed using exponents, is there a way to capitalize on that? Does the fact that $29$ is raised to the power of $17$ help in calculating the LCM or GCD? Ecuclidean algorithm requires me to first solve $29^{17}$ and then use the numbers obtained to calculate the GCD – Ray Penber Oct 24 '19 at 08:10
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    The Euclidean algorithm requires no such thing. It just requires you to calculate $29^{17}+2-(29^{17}-1)$. – Arthur Oct 24 '19 at 08:12
  • @Arthur I'm surprised you got the GCD by just subtracting both the numbers. Can u explain a little? I thought Euclidean algo was like say u wanted LCM(30,27) , you would write $30=1\cdot27 + 3$ and then proceed with $27$ and $3$ and on. – Ray Penber Oct 24 '19 at 08:18
  • Move $1\cdot 27$ over to the other side of the equality sign, and you will have $30-27 = 3$. That's the subtraction I'm referring to. Now do it with your two numbers instead of $30$ and $27$. – Arthur Oct 24 '19 at 08:20
  • Oh ok. I'll try to first find the GCD and then solve for LCM. – Ray Penber Oct 24 '19 at 08:24

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Let do this: $$(29^{17}+2,29^{17}-1) = (29^{17}+2 - 29^{17}+1, 29^{17}-1) = (3,29^{17}-1)$$ $$29^{17}-1 \overset{3}{\equiv} (-1)^{17}-1 \overset{3}{\equiv} -1-1 \overset{3}{\equiv} -2$$ $$\Longrightarrow (29^{17}+2,29^{17}-1) = (3,29^{17}-1) = (3,-2) = 1$$ Now you can continue your way.

Ali Ashja'
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Hint $\, \gcd(a\!+\!3,a) =\!\!\!\!\overbrace{ \gcd(3,a)}^{\large \gcd(\color{#c00}{3,\,29^{17}-1})}\!\!\! $ by subtractive Euclid algorithm (subtract least from largest arg)

Finally note that: $\bmod \color{#c00}{3}\!:\ \underbrace{\color{#c00}{29^{17}\!-\!1}\equiv (-1)^{17}\!-1}_{\textstyle{29\ \equiv\ -1\ \ }}\equiv -2\not\equiv 0\ $ by the Congruence Power Rule.

Bill Dubuque
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