Question

Show that p2 will leave a remainder 1 when divided by 8 if p is an odd positive integer

Answer 1

Let p be any odd positive integet

p = 2n + 1

p² = (2n + 1)² = 4n² + 4n + 1

p² = 4n(n + 1) + 1

for n = 0. p² = 1

p²/8 remainder = 1

for n = 1

p² = 9 . p² / 8 remainder = 1

hence we can see for any value of n p²/8 remainder = 1

So any odd positive number divided by 8 gives remainder 1

Answer 2

Explanation:

p = 2n + 1 where n = 0, 1, 2, 3, 4, ...

=> p² = 4n² + 4n + 1

= 4(n² + n) + 1

= 4n(n+1) + 1

but n(n+1) is even because if n is odd, n+1 is even - otherwise n is even

therefore 4n(n+1) + 1 = 8m + 1 where m is an integer.

=> p² = 1

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