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summer camp blue group number theory day 1 P9

 
 
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Re: summer camp blue group number theory day 1 P9
by Areteem Professor - Wednesday, August 26, 2020, 12:07 PM
 

Yes! That's the idea. Since each of $\sum a_k$ and $\sum b_k$ are $\dfrac{m}{2} \pmod{m}$, we have $\sum a_k + b_k = 0 \pmod{m}$, but we needed $\sum a_k + b_k = \dfrac{m}{2} \pmod{m}$.