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

Jolb
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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
 
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Hint: compare what you have with what you need.
 
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.
 
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