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} > x^{2} + 1

for

x

as an integer.

If this were true, then $\sqrt{x^{2} + 1}$

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 $x = \sqrt{m}$ we need

$(x + 0.0001)^{2} > x^{2} + 1$

$x^{2} + 0.0002 x + 0 . 0001^{2} > x^{2} + 1$

$0.0002 x > 1 - 0 . 0001^{2}$

$x > 5000 - (1 / 20000)$

$m > 24999999.5$

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 = \sqrt{m - 1}

Then we require

$\sqrt{m} < x + 0.0001$

$m < (x + 0.0001)^{2}$

$x^{2} + 1 < (x + 0.0001)^{2}$

Steve

Yatir Halevi

Posted on Tuesday, 28 January, 2003 - 04:34 pm:

Thanks.
I get it now!