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?
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$\begingroup$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.
$\endgroup$ $\begingroup$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.
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