Hay I am going over some old exams and hit this:
(a) Use the Euclidean algorithm to show that $\gcd(60; 17) = 1$.
(b) Hence find integers $x, y$ satisfying $60x + 17y = 1$.
(c) Find another solution in integers to $60x + 17y = 1$ (distinct from your previous answer).
Finding the gcd is easy but i can seem to find any help on the rest. Can someone point me at something that might help?
For (a), you should have seen that $$60=17\cdot 3+9\\17=9\cdot 1+8\\9=8\cdot 1+1,$$ whence $$\begin{align}1 &= 9-8\\ &= 9-(17-9)\\ &= 9\cdot 2-17\\ &= (60-17\cdot 3)\cdot 2-17\\ &= 60\cdot 2+17\cdot(-7).\end{align}$$ For (c), you'll need to pick an $a$ and a $b$ such that $a,b\ne 0$ and $$1=60\cdot(1+a)+17\cdot(-7+b).$$ I leave it to you to find such $a,b$. It isn't too hard.
This is the time to review the Extended Euclicean Algorithm. This algorithm uncovers $d = \gcd(a, b)$ as well as the $x, y$, in: $ax + by = d$.
You may choose among many ways to determine $x, y$; each method you find at the link above demonstrates the method with an example. First walk through the given examples, then apply to your question, going back and forth, as needed, to learn a method well.
But do learn one method/strategy, and learn the the method well; don't just let others' hand you an answer: you won't learn that way. You need to know a method well enough to use it and apply it when preparing for and taking an exam, e.g. And try testing your skills when the $\gcd(a, b) \neq 1$.
So it's important that you realize how these parts fit together. Let's do the first bit, i.e., go through the Euclidean algorithm: $$ 60 = 3(17)+9\\ 17=9+8\\ 9=8+1 $$ Since we've reached a remainder of $1$, we can conclude that the gcd of the 2 numbers is, in fact, $1$. In order to solve the second part, we can now use our work, reversing our steps as follows: $$ 1=9-8\\ 1=9-(17-9)=2(9)-17\\ 1=2(60-3(17))-17=2(60)-7(17) $$ That is, we now have the solution $x=2,y=-7$. To get the other answers, add $17\mu$ to $x$ and subtract $60\mu$ from $y$ for some integer $\mu$.
