maths help

Question

prove that none of the integers 11,111,1111,11111,... is a square of an integer

Answer 1

 

11, 111, 1111, 11111, 111111,...

prove none of these numbers are perfect squares

 

an integer is either even or odd

let x be an even integer

x=2n is the form of an even integer

square this even integer

x²=4n² which is a multiple of 4 (the square of any even integer is a multiple of 4)

for a number to be a multiple of 4, the number formed by the two right-most digits must be divisible by 4

none of the numbers listed is divisible by 4 because in all of the numbers the two right-most digits form the

number 11 which is not divisible by 4

therefore none of the numbers listed is the square of an even integer

 

let x be an odd integer

x=2n+1 is the form of an odd integer

square this odd integer

x²=4n²+4n+1 is one more than a multiple of 4  (4n²+4n is a multiple of 4)

if any of the numbers listed is the square of an odd integer, than it must be one more than a multiple of 4

subtract 1 from any of the numbers listed and the two right-most digits form the number 10 which is not divisible by 4

therefore none of the numbers listed is the square of an odd integer

therefore none of the numbers listed are squares of any integers, even or odd

Answer 2

A fairly simple way to see that this is true for an even number of 1's is to note that

 

9x 1111 =  10^4 - 1   ,   so  1111 = (10^4 -1 )/9

this number cannot be a perfect square because 9 is a perfect square but 10^4 -1 is one less than a perfect square and so is not a perfect square.

 

This generalizes.    The number with  m  1's can be written as   (10^m - 1)/9.

 

If m is even, the same argument is the same as for 1111

 

It seems likely that 10^m -1  is cannot be a perfect square for odd m also.    However, I don't see a proof this morning.