Can somebody help me prove that there exists a solution in the whole numbers for $5^{200}x+7^{500}y=1$

Question

ivancsc1996 · Answer

In other words $\exists x,y \in \mathbb{Z}:5^{200}x+7^{500}y=1$

perl · Answer

Is the equation
$ \large 5^{200}\cdot x+7^{500}\cdot y=1$

perl · Answer

The diophantine equation ax+by=c has solutions if and only if gcd(a,b)|c

does gcd( 5^200, 7^500) | 1 ?

perl · Answer

typo
The diophantine equation ax+by=c has `integer` solutions if and only if gcd(a,b)|c.

ivancsc1996 · Answer

@perl We haven't proven that in class. I need to know how to prove that $gcd(a,b)=1\rightarrow \exists x,y \in\mathbb{Z}:ax+by=1$

perl · Answer

this is a good resource for a proof http://math.stackexchange.com/questions/20717/how-to-find-solutions-of-linear-diophantine-ax-by-c

perl · Answer

@mathmath333

Why does taking the mod 5 of both sides prove there is a solution to the original equation. 
Are you constructing the general solution to this problem?
Also can you explain how you got x = 5n, if y = 1 mod 5  In your proof 
It looks like the value of x did not matter in the argument , because 5^200 x mod 5 = 0

mathmath333 · Answer

i probably misread the question, i thought he asked for a solution of question

mathmath333 · Answer

in that the gcd idea work fine.