Ngoc Tran
Posted on Monday, 27 January, 2003 - 02:04 pm:
find the smallest positive integer m such that m is not a square and yet in the decimal expansion of sqrt(m), the decimal point is followed by at least 4 consecutives zeros.

Thanks for your help,
Hugh Venables
Posted on Monday, 27 January, 2003 - 02:28 pm:
A idea to start with is that this number must be one larger than a perfect square (which you could call x^2) that is large enough so that adding 1 to it (to get x^2+1 - the number you want) only adds 0.0001 to x (well, just less but there will be rounding as dealing with integers).
Yatir Halevi
Posted on Monday, 27 January, 2003 - 02:33 pm:
I checked up to 5 million and didn't find any. Does this number exist?

Yatir
Stephen Burgess
Posted on Monday, 27 January, 2003 - 02:48 pm:
Consider (x+.0001)2 > x2 +1 for x as an integer.

If this were true, then
  ____
Öx2+1
 

Hugh Venables
Posted on Tuesday, 28 January, 2003 - 11:50 am:
The number does exist and can be worked out using a formula (and it's bigger than 5 million)
Yatir Halevi
Posted on Tuesday, 28 January, 2003 - 12:40 pm:
The number is 25000001
Ngoc Tran
Posted on Tuesday, 28 January, 2003 - 12:42 pm:
How did you work it out?
Yatir Halevi
Posted on Tuesday, 28 January, 2003 - 12:47 pm:
QBASIC
Steve Megson
Posted on Tuesday, 28 January, 2003 - 01:58 pm:
That's one way.

As Stephen says above, for x=Öm we need

(x+0.0001)2 > x2 +1

x2+0.0002x+0.00012 > x2 + 1

0.0002 x > 1-0.00012

x > 5000 - (1/20000)

m > 24999999.5

So any ''one more than a perfect square'' above 25000000 works, and 25000001 is the least of these.

Steve

Yatir Halevi
Posted on Tuesday, 28 January, 2003 - 02:53 pm:
I don't see why the formula works....
Steve Megson
Posted on Tuesday, 28 January, 2003 - 04:19 pm:
Correction: that should be
x=   ___
Öm-1
 

.

Then we require

Öm < x+0.0001

m < (x+0.0001)2

x2+1 < (x+0.0001)2

Steve

Yatir Halevi
Posted on Tuesday, 28 January, 2003 - 04:34 pm:
Thanks.
I get it now!