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        riddles >> medium >> number 40....09 not a perfect square
        
(Message started by: gkwal on Jul 16^th, 2007, 9:28pm)

Title: number 40....09 not a perfect square
Post by gkwal on Jul 16^th, 2007, 9:28pm

Prove that the number 40..09 (with at least one zero) is not a perfect square.

Title: Re: number 40....09 not a perfect square
Post by towr on Jul 17^th, 2007, 1:29am

For an odd number of 0's it's easy [hide]then 40..09 - 9 is a square and (sqrt(40..09-9)+1)² will be much larger[/hide].

Title: Re: number 40....09 not a perfect square
Post by Eigenray on Jul 18^th, 2007, 12:13am

Suppose that 4*10^k + 9 = x². That is,[hideb]4*10^k = x²-9 = (x+3)(x-3).

Let A=x+3, B=x-3, and let d = gcd(A,B). Since A - B = 6, we must have d | 6. But since AB = 4*10^k is not divisible by 3, and A,B are both even (x is odd), we can only have d=2.

Now, we have AB = 2^k+25^k, with gcd(A,B)=2, and A>B. There are only two ways this can happen:

(1) A = 2^k+15^k, B=2. But then

6 = A - B = 2(10^k-1),

which is impossible.

(2) A = 2*5^k, B = 2^k+1. Now

6 = A - B = 2(5^k - 2^k),

or 5^k - 2^k = 3, which happens only for k=1. But this is just the case 49 = 7².

So there are no other solutions.[/hideb]

Title: Re: number 40....09 not a perfect square
Post by Barukh on Jul 18^th, 2007, 2:07am

I love your solution, Eigenray! :D