SOLUTION: Prove that if n is any integer not divisible by 5, then n has a square that is either of the form 5k+1 or 5k+4, where k is some integer.

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Question 755184: Prove that if n is any integer not divisible by 5, then n has a square that is either of the form 5k+1 or 5k+4, where k is some integer.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
If n is any integer not divisible by 5, when we divide it by 5 we get 1, 2, 3, or 4 as a remainder. That means that
n=5p%2B1 (remainder=1), or
n=5p%2B2 (remainder=2), or
n=5p%2B3 (remainder=3), or
n=5p%2B4 (remainder=4),
for some integer p.

If n=5p%2B1, then
n%5E2=%285p%2B1%29%5E2=25p%5E2%2B10p%2B1=5%285p%5E2%2B2p%29%2B1,
which is n%5E2=5k%2B1 with k=5p%5E2%2B2p
If n=5p%2B2, then
n%5E2=%285p%2B2%29%5E2=25p%5E2%2B20p%2B4=5%285p%5E2%2B4p%29%2B4,
which is n%5E2=5k%2B4 with k=5p%5E2%2B4p
If n=5p%2B3, then
n%5E2=%285p%2B3%29%5E2=25p%5E2%2B30p%2B9=25p%5E2%2B30p%2B5%2B4=5%285p%5E2%2B6p%2B1%29%2B4,
which is n%5E2=5k%2B4 with k=5p%5E2%2B6p%2B1
If n=5p%2B4, then
n%5E2=%285p%2B4%29%5E2=25p%5E2%2B40p%2B16=25p%5E2%2B40p%2B15%2B1=5%285p%5E2%2B8p%2B3%29%2B1,
which is n%5E2=5k%2B1 with k=5p%5E2%2B8p%2B3