Thread: (Perfect) Squares

1.) Prove that a square of some integer has to leave a remainder of either 0 or 1 when it is divided by 4.


2.) Prove that there is no number in the following sequence: 11, 111, 1111, 11111, ... that's a perfect square.

1. Let the number you're after be p. There are only 4 possible senarios in modulo 4.

p = 0 (mod 4) => p^2 = 0 (mod 4)
p = 1 (mod 4) => p^2 = 1 (mod 4)
p = 2 (mod 4) => p^2 = 4 = 0 (mod 4)
p = 3 (mod 4) => p^2 = 9 = 1 (mod 4)

Therefore any square number would have the property that p = 0 or 1 (mod 4)


2. As you can see, you can write each of the 1111.....11 (n digits) numbers as:

(n-2) digits ---> 111...111 x 100 + 11

In mod 4, this is the same as 111...111 x (4 x 25) + 11 = 11 = 3 (mod 4)

Since 111...111 is not equal to 0 or 1 mod 4, it cannot be a perfect square (from 1)

Originally Posted by Ideasman

2.) Prove that there is no number in the following sequence: 11, 111, 1111, 11111, ... that's a perfect square.

This one is simple.

Because,
111..1111 = 111....108 +3

But 111....108 is divisible by 4 for the last two digits are.
Thus, it is expressible as 4k.

Thus,
111....1111=4k+3 for some k in Z.

But by the previous discussion squares are of the form 4k or 4k+1. Impossible.