Multiple of prime p + multiple of (integer<p) = 1 proof?

[Jolb]

Let p be a prime number and x be some positive integer less than p.

How do I prove that there exist integers a and b such that
1 = ax + bp

 

[dodo]

Hint: compare what you have with what you need.

 

[robert Ihnot]

Since any term less than p is relatively prime to p, and thus has a solution of 1; it is easy to choose a desireable case.