prove to disprove:
a)If there exists an integer solution to ax^2+by^2=c, then gcd(a,b)|c
b)if gcd(a,b)|c, there exists an integer solution to ax^2+by^2=c.
I don't know where to start..any help please.. thank you~

Question

prove to disprove: 
a)If there exists an integer solution to ax^2+by^2=c, then gcd(a,b)|c
b)if gcd(a,b)|c, there exists an integer solution to ax^2+by^2=c.
I don't know where to start..any help please.. thank you~

anonymous · Answer

a) first part is clear, suppose there exists an integer solution to ax^2+by^2=c and$$d=\gcd(a,b)$$then$$d|ax^2+by^2=c \ \ \rightarrow \ \ d|c$$

anonymous · Answer

b) second part, let a=1, b=2 and c=3...see what happens