Hint Suppose r1 r2 Consider ac ab acbSolutionSupposer r1r

Hint: Suppose r1 = r2. Consider ac - ab = a(c-b).

Solution

Supposer r1=r2

Hence, r1=ab=r2=ac %p

ie ab=ac % p

a(c-b)=0 %p

Hence, a(c-b) is a multiple of p

But, 1<=a<=p-1
So, p cannot divide a

Hence, p|(c-b)

But, 1<=b<c<p

So, 0<c-b<p

Hence, p |c-b is not possible

Hence a contradiction

Hence, r1 and r2 are not equal

Hint: Suppose r1 = r2. Consider ac - ab = a(c-b).SolutionSupposer r1=r2 Hence, r1=ab=r2=ac %p ie ab=ac % p a(c-b)=0 %p Hence, a(c-b) is a multiple of p But, 1&l

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