MathLabs

Problem 5

Let a,b,ca,b,c be integers satisfying 0<a<c−10<a<c-1 and 1<b<c1<b<c. For each kk, 0≤k≤a0\le k\le a, let rkr_k, 0≤rk<c0\le r_k<c, be the remainder of kbkb when divided by cc. Prove that the two sets {r0,r1,r2,…,ra}\{r_0,r_1,r_2,\ldots,r_a\} and {0,1,2,…,a}\{0,1,2,\ldots,a\} are different.
Step 5 of 5: Reach the contradiction
b≢1,a+b+1(modc)⟹b≡ab+b(modc)⟹a≡0(modc)b\not\equiv1,a+b+1\pmod c\Longrightarrow b\equiv ab+b\pmod c\Longrightarrow a\equiv0\pmod c
Detailed analysis

In the congruent multisets, b cannot match 1 because 1<b<c, and cannot match a+b+1 because 0<a+1<c. Thus b≡ab+b modulo c. Since gcd(b,c)=1, this implies a≡0 modulo c, contradicting 0<a<c−1. Hence the residue sets are different.