2.5.22) Note that the second line in the given solution is wrong! 
 
 Here is a correct solution:  

 Calim 1: If  d=(c,n), then 1=(c/d,n/d).

Since d=(c,n), we have d|c, d|n and d=cx+ny for some x, y in Z. 

Hence we can write c=dr, n=dm for some r and m in Z.

Then d=drx+dmy=d(rx+my). Therefore by the cancellation law in Z we

obtain 1=rx+my. Then 1=(r,m)=(c/d,n/d) as claimed.


Since ca is conruent to cb modulo n, c(a-b)=nk for some k in Z.

Let c=dr, n=dm. Then dr(a-b)=dmk, and by the cancellation law in Z

we have r(a-b)=mk. Hence m|r(a-b).

On the other hand by Claim 1, 1=(r,m). 

Then by Theorem 2.14 it must be the case that m|(a-b).
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2.26. 20) In the third line from the bottom of the solution given 

it should be mentioned that product of non-zero elements cannot be zero in Zp

because of Excercise 2.6. 18.


Here is an alternative solution (without using Exc. 2.6.18):

suppose x is in Zp and it has its own inverse.  

Then xx is congruent to 1 in Zp.

Hence xx-1=(x-1)(x+1) is conruent to zero in Zp.

Then p|(x-1)(x+1). Since p is a prime, either p|(x-1)  or p|(x+1).

If p|(x-1),  then x is congruent to 1 in Zp.

If p|(x+1), then x is conruent to -1 in Zp.

Since -1 is congruent to p-1, 

transitivity of congruence implies that x is congruent to p-1.