Question

Show that P² will leave a remainder 1 when divided by 5 for any positive integer P

Answer 1

Step-by-step explanation:

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 = 1p²/8 remainder = 1

for n = 1

p² = 9 . p² / 5 remainder = 1

hence we can see for any value of n p²/5 remainder = 1

So any odd positive number divided by 5 gives remainder 1

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Answer 2

To Prove:- we had to show that any positive integer divided by 5 will give the remainder as 1

It is given that p = 2n+1

p = 2n + 1

If, p = 2n+1 then, p² = (2n+1)²

p² = (2n + 1)²

p² = 4n² + 4n + 1 ((a+b)² = a² + 2ab + b²) formula)

Let us take 4n as common then,

p² = 4n(n + 1) + 1

If, n = 0. p² = 1p²/8  = 1

Then, n = 1

If n = 1 then,

p² = 9*p² / 5 = 1

Therefore, p = 9*p² /5 = 1

So we had showed that any positive integer divided by 5 will give the remainder as 1