Prove that either 2.10^500 +15 or 2.10^500 +16 is not a perfect square.Is your proof constructive or nonconstr

Question

Use methods of Proof

Answer 1

(2.10^500)+15 has last two digits 15. It cannot be a perfect square as a perfect square if last digit is 5 it has to be 25.

proof:
last digit is 5 means it is divisible by 5.

(10x+5)^2 = 100x^2+100x + 25 , so last 2 digits 25

2.10^500+16 has last 2 digits 16.

(10x+4)^2 = 100x^2+80x + 16
(10x-4)^2 = 100x^2-80x+ 16
if last 2 digits are 16 then both the previous digit cannot be zero

Answer 2

Do you mean 'a perfect square of integers'?

Proof by contradiction:

Assume that 2.1^500 + 15 is a perfect square. Therefore, it is a positive integer. Therefore 2.1^500 is a an integer. But, this is false because the binomial expansion of (2 + .1)^500 will take the form of:

2^500 + 500*2^499*(.1) + 500 Choose 2 * 2^498 * (.1)^2 + ... +
(.1)^500.

As there will only be one term in this expansion that has a non-zero digit in the 500th decimal place, it cannot add to something else to clear the 500th decimal place. Therefore, it cannot be an integer as it must have a 1 in the 500th decimal place.

Therefore, the assumption that 2.10^500 + 15 must be false.

It follows by the same logic that 2.10^500 + 16 cannot be a perfect square, either.

This is a non-constructive method.

Answer 3

2*10^500 + 15 cannot be a square, because it ends in 15.
No square ends in 15. (Look at the remainders of
all squares when you divide by 100.)
Constructive proof(?).

Answer 4

Do I get money for this? Math is not my bag. :)

Answer 5

ccc