Integer square of the form 111...1?

Question

Can an integer of the form 111...1 (there are n digits, with n ≥ 2) be the square of an integer?

Answer 1

1^2 = 1

11^2 = 121 which is close to 111

11 is close to 9 =3^2

perfect squares are

0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 169, ...441, ...

0 1^2, ...9^2,....11^2....19^2,..21^2 . 29^2, ...31^2, ...111^2, ...

for 111...1 to be a perfect square it would have to be the square of an odd number ending in 1 or 9, 3,5,7 won't work.

to be a perfect square it must divide by 4 with a remainder of 0 or 1

but 11/4 =2 R=3, 111/4 = 27 R=3, 1111/4 = 277 R=3, 111...1/4 always leave a Remainder of 3, never 0 or 1, Except for 1/4 = 0 R=1, so 1 is a perfect square

Answer 2

A number consisting of n≥2 digits of 1s, such as 11, 111, or 1111, cannot be a perfect square. These numbers, known as repunits, can be expressed as (10ⁿ-1)/9. For n≥2, all such numbers are congruent to 3 modulo 4, meaning they leave a remainder of 3 when divided by 4. However, the square of any integer is always congruent to either 0 or 1 modulo 4. Since 3 modulo 4 is not a possible result for a perfect square, no such repunit with n≥2 digits can be a perfect square.

Answer 3

The answer is NO. An integer of the form 111...1 is an odd number. If it were a perfect square of another integer, that integer must be odd, and of the form 2k+1, where k is an integer. Then its square (2k + 1)² = 4k(k + 1) +1 has remainder 1 when divided by 4, whereas a number of the form 111...1 is not. Indeed, if this number has more than 2 digits, then it is of the form

111...100 + 11, which is the sum of a multiple of 100 and 11. Now 100 is a multiple of 4, but 11 has remainder 3 when divided by 4, so it cannot be the square of an odd integer. If this number has only two digits, then it is just 11.