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The problem is the following: For each $a \in \Bbb{Z}$ work out $\gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$

This is what I have at the moment:

Let's call $d = \gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$

Then $d \ \vert \ 3^{16} \cdot 2a + 10 \ \land d \ \vert \ 3^{17} \cdot a + 66$

$\Rightarrow d \ \vert \ (3^{16} \cdot 2a + 10) \cdot 3 \ \land \ d \ \vert \ (3^{17} \cdot a + 66) \cdot 2$

$\Rightarrow d \ \vert \ 3^{16} \cdot 6a + 30 \ \land \ d \ \vert \ 3^{16} \cdot 6a + 132$

$\Rightarrow d \ \vert \ 3^{16} \cdot 6a + 132 - (3^{16} \cdot 6a + 30) \ = \ 102 \ = \ 2 \cdot 3 \cdot 17$

Also, $d \ \vert \ (3^{16} \cdot 2a + 10) \cdot 33 \ \land \ d \ \vert \ (3^{17} \cdot a + 66) \cdot 5$

$\Rightarrow d \ \vert \ 3^{17} \cdot 17a$ (almost with the same method as before)

So I get $d \ \vert \ 102 \ \land d \ \vert \ 3^{17} \cdot 17a$

After this, I can't see how to continue.

Davood
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tanoserio
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4 Answers4

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The prime factorization of $102$ is $2 \times 3 \times 17$. When do $2$, $3$ and $17$ divide your numbers?

Robert Israel
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From $d|102$ you have $d\in\{1,2,3,17,34,51,102\}$ and because $d|3^{16}2a+10$ you have $d\notin \{3,51,102\}$. Clearly $2|d$ iff $2|a$. And with $17$ you have a lot of different subcases to deal with. If $17|a$ then $d\ne 17$...

nonuser
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Hint $\ \ \ (d,102) = (d,\,2\cdot 3\cdot 17) = (d,2)\,(d,3)\,(d,17).\, $ Computing these gcds by Euclid & Fermat

$\qquad\quad\ \begin{align} (d,2) &= (\color{#c00}{3^{16}(2a)\! +\! 10}, \ \color{#0a0}{3^{17} a\! +\! 66},\,2) = (\color{#c00}0,\color{#0a0}a,2) = (a,2)\\[.2em] (d,3) &= (\color{#c00}1,\ \color{#0a0}0,\ 3) = 1,\ \text{ and, finally, using $\,\color{#c00}{3^{16}}\equiv 1\!\!\!\!\pmod{\!17}\ $ we have}\\[.2em] (d,17) &= (\color{#c00}{2a\!+\!10},\,\color{#0a0}{3a\!-\!2},17) = (2a\!+\!10,a\!+\!5,17) = (0,a\!+\!5,17) \end{align}$

Hence $\, (d,102) = (a,2)\,(a\!+\!5,17)$

Remark $\, $ The elimination that you employed to deduce that $\,d\mid 102\,$ is a special case of the determinant criterion presented here. Namely, applying that with $\,b=3^{16}\,$ yields

$\ (ab,1)\mapsto (2ab\!+\!10,3ab\!+\!66)\,$ has $\,\det = 2(66)\!-\!10(3) = 102\ $ so $\ d\mid 102(ab,1) = 102$

Bill Dubuque
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  • Can you explain how do you reach the a in (0, a, 2) when you try to produce (d, 2)? – tanoserio Sep 20 '17 at 22:07
  • @est By Euclid's algorithm, if one gcd argument is $2$ then we can mod out all the other gcd args by $2,,$ so $\bmod 2!:\ \color{#0a0}{3^{17}a+66 \equiv 1\cdot a + 0\equiv a}\ $ by $,66\equiv 0,$ and $,3^{17}\equiv 1^{17}\equiv 1,,$ by $,3\equiv 1.\ $ Similarly we mod out the other gcd args by $3$ in $,(d,3),,$ and by $17$ in $,(d,17),$ – Bill Dubuque Sep 20 '17 at 22:29
  • @est Generally $\ (a,b,c,\ldots) = (a, b\bmod a,, c\bmod a,\ldots).\ $ This is the reduction (descent) step in the Euclidean algorithm for the GCD – Bill Dubuque Sep 20 '17 at 22:33
  • You are right. Last question, when you do in the (d, 17), I arrived at $gcd(2a + 10, 3a + 15, 17)$ which is equal to what you wrote but then how do you apply Euclid's? – tanoserio Sep 20 '17 at 23:02
  • I think I've got it. $(2a + 10, 3a + 15, 17) = (2(a + 5), 3(a + 5), 17)$ and I think this is equal to $ (0, a + 5, 17) $. Is this wrong? – tanoserio Sep 20 '17 at 23:27
  • @est I'd need to know how you did it. Here's one way. Writing $,a_i$ for the $i$'th gcd argument, subtracting $a_1$ from $a_2$ yields $(2a+10,a+5,17),,$ then subtracting $,2a_2$ from $a_1$ yields $(0,a+5,17) = (a+5,17).,$ This uses the GCD property $,(a,b) = (a,b!+!na),,$ a generalization of Euclid's $,(a,b) = (a,b\bmod a).\ \ $ – Bill Dubuque Sep 20 '17 at 23:38
  • didn't remember that property. I used another one that says that $ gcd(m·c, m·b) = m·gcd(c, b)$ with $ m $ being $ (a + 5) $ and since $ gcd(2, 3) = 1 \ \Rightarrow \ m·gcd(c, b) = a + 5 $. But both ways I think are right. – tanoserio Sep 20 '17 at 23:43
  • @est Yes, that's correct. You used the GCD Distributive Law. – Bill Dubuque Sep 20 '17 at 23:56
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$$E=\gcd(3^{16}2a+10,3^{17}a+66)=\gcd(b,c)=\gcd(b,c,3c-2b)=\gcd(b,c,102)$$ $$E=\gcd(b \mod 102,b\mod 102)=\gcd(36 a+10,3a+66)=\gcd(d,e)$$ $$E=\gcd(d-12e \mod 102,e)=\gcd(34,e)=\gcd(34,3(a+22))=\gcd(34,a+22)$$ because $34 \mod 3 \neq 0$

so $E=\gcd(a,2)\times \gcd(a+5,17)$

Dattier
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