Question

Q. Show that a perfect square never leaves remainder 2 on dividing by 3.​

Please let my i d answer first :) ​

Answer 1

Answer:

Thank you for letting me answer

Please refer the attachment for the answer.

Hope the answer helps you

Answer 2

x=0mod3 + x = 3n →

x² = 9n² = 3(3n²) == 3m,

x=1mod3 + x = 3n+1 →

x² = 9n²+6n+1 = 3(3n²+2n)+1= 3m+1,

x=2mod3 x = 3n+2 →

x² = 9n²+12n+4= 3(3n²+4n+1)+1==3m+1;

so f(x)=x² sends:

00mod3, 11mod3, 2-1 mod3