Uniqueness of integers question

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BustedBreaks
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Find integers s and t such that 1 = 7*s + 11*t. Show that s and t are not unique.

I can find numbers that satisfy this question, t=2, s=-3 and t=-5, s=8, that show s and t are not unique, but this doesn't seem to be rigorous and I'm not sure where to start with proving this.
 
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In general there is a theorem which states that for any two relatively prime positive integers a and b there exist integers x and y such that: 1=ax+by. Or, a and b are relatively prime iff there exist integers x and y such that 1=ax+by.
 
sutupidmath said:
In general there is a theorem which states that for any two relatively prime positive integers a and b there exist integers x and y such that: 1=ax+by. Or, a and b are relatively prime iff there exist integers x and y such that 1=ax+by.

yea.. or there exist integers x and y such that (a,b)=ax+by.
 
icystrike said:
yea.. or there exist integers x and y such that (a,b)=ax+by.

THis is an even more general result. Namely if L={n|n=ax+by, x,y in Z}, then the smallest element of L is gcd(a,b).
 
BustedBreaks said:
Find integers s and t such that 1 = 7*s + 11*t. Show that s and t are not unique.

I can find numbers that satisfy this question, t=2, s=-3 and t=-5, s=8, that show s and t are not unique, but this doesn't seem to be rigorous and I'm not sure where to start with proving this.
But this is completely correct and a perfectly good solution to this problem. 7(-3)+ 11(2)= -21+ 22= 1, 7(8)+ 11(-5)= 56- 55= 1, and these are two distinct solutions. That is all that is necessary to show that the solution to this particular problem is not unique.