Prove (15n+17,10n+11) = 1

Question

I am new to number theory and wanted to know If I am doing this correctly:

W.t.s. $(15n+17,10n+11) = 1$

Using the Division Algorithm we have

$15n + 17 = (10n + 11)(1) + (5n + 6)$

$10n + 11 = (5n + 6)(1) + (5n + 5)$

$5n + 6 = (5n + 5) + 1$

$\Rightarrow$ $(15n + 17,10n + 11)$ = ($5n + 5,1)$, but $(5n + 5,1)$ = $1$ (1 is relatively prime to every integer)

$\therefore$ $(15n+17,10n+11) = 1$ as needed?

@AlexS He is using the euclidean algorithm and I assume he meant $(5n+5,1)=1$. Other than that it looks correct to me. — CyclotomicField, Feb 02 '19 at 02:30
@AlexS lol yeah I meant $(5n+5,1) = 1$ , division algorithm. — homosapien, Feb 02 '19 at 02:31
@Hossien Sahebjame That's what I thought, but wanted to be sure. And yes, your solution is correct, because the algorithm halts and returns a value of 1. — Alex S, Feb 02 '19 at 02:36
@AlexS Awesome! thought that too, wanted to make sure. Thanks so much!! — homosapien, Feb 02 '19 at 02:37

score 3 · Accepted Answer · answered Feb 02 '19 at 02:29

3

Eliminate $n$

$$2(15n+17)-3(10n+11)=?$$

answered Feb 02 '19 at 02:29

lab bhattacharjee

Im confused how do I use this to show it holds $\forall n$? Isn't this just one particular ($x_0,y_0)$ satisfying the diophantine equation since we know it is solvable? ($a = 15n + 17$ and $b = 10n + 11$) – homosapien Feb 02 '19 at 02:35
1

@Hossien Sahebjame The g.c.d. of a and b is a $\mathbb{Z}$ linear combination of $a$ and $b$. The value of $n$ is eliminated regardless of the choice of its value ($215n-310n=0$, $\forall n\in\mathbb{Z}$). – Alex S Feb 02 '19 at 02:41

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