n^2 is a multiple of 100

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Question from Rahul:

I have to prove that n^2 is a multiple of 100 is necessary or
Sufficient condition (or both) for n being multiple of 10.
I can see that if n=10*p for some p then n^2 = 100*p^2, and p^2 is
also an integer as p is.
n^2= 100*p => n=10* sqrt(p) and here sqrt(p) need not be integer.

I still feel that n^2 = 100*p will imply n=10*p for two different p's.

Also I can not see any immediate counter example to prove that
n^2=100*p for p integer, does not imply n = 10*p or q etc.

Hi Rahul,

I can give you a hint. $100 = 4 \times 25.$

Write back if you need more assistance.

Penny

Rahul wrote back

I had been given a hint to the problem that 100 = 4*25.
I can say 4*25 = 10*10 which is multiple of 10, what next?

Rahul,

You are assuming that $n^2 = 100 \times p$ and thus $n^2 = 4 \times 25 \times p.$ Therefore $4$ divides the right side and hence it divides the left side. Hence $4 = 2^2$ divides $n^{2}.$ Can you complete the problem now?

Penny

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