This article is from the Puzzles FAQ, by Chris Cole chris@questrel.questrel.com and Matthew Daly mwdaly@pobox.com with numerous contributions by others.

Title: Cliff Puzzle 8: Squares and Squares and Squares ....

From: cliff@watson.ibm.com

If you respond to this puzzle, if possible please send me your name,

address, affiliation, e-mail address, so I can properly credit you if

you provide unique information. PLEASE ALSO directly mail me a copy of

your response in addition to any responding you do in the newsgroup. I

will assume it is OK to describe your answer in any article or

publication I may write in the future, with attribution to you, unless

you state otherwise. Thanks, Cliff Pickover

* * *

1. What is the smallest square with leading digit 1 which remains a

square when the leading 1 is replaced by a 2?

In other words, if x**2 = 1.........., is there a y**2 = 2......... ?

2. What is the smallest square with leading digit 1 which remains a

square when the leading 1 is replaced by a 2 and also remains a square

when the leading digit is replaced by a 3?

3. What is the smallest square with leading digit 1 which remains a

square when the leading 1 is replaced by a 2, and also remains a square

when the leading digit is replaced by a 3, and also remains a square

when the leading digit is replaced by a 4?

pickover/pickover.08.s

-------------------------

> 1. What is the smallest square with leading digit 1 which remains a

> square when the leading 1 is replaced by a 2?

>

11025 ( 105 * 105 ) ---- 21025 ( 145 * 145 )

>

> 2. What is the smallest square with leading digit 1 which remains a

> square when the leading 1 is replaced by a 2 and also remains a square

> when the leading digit is replaced by a 3?

>

No solution till 1,000,000,000.

> 3. What is the smallest square with leading digit 1 which remains a

> square when the leading 1 is replaced by a 2, and also remains a square

> when the leading digit is replaced by a 3, and also remains a square

> when the leading digit is replaced by a 4?

>

>

No solution till 1,000,000,000.

The property that you are looking for ( however with different leading

digits ) is owned by the following numbers.

2025 3025 ------------- 11025 21025 57600 67600 --------------- 202500 302500 342225 442225 ------------------ 1102500 2102500 3515625 4515625 5760000 6760000 ------------------- 11390625 21390625 20250000 30250000 34222500 44222500 ---------------------- 110250000 210250000 196700625 296700625 351562500 451562500 576000000 676000000 -------------------------

This is probably of no use to you, but, anyway.

-------------------------

In article <1992Oct20.184149.51596@watson.ibm.com> you write:

>Title: Cliff Puzzle 8: Squares and Squares and Squares ....

>1. What is the smallest square with leading digit 1 which remains a

>square when the leading 1 is replaced by a 2?

>In other words, if x**2 = 1.........., is there a y**2 = 2......... ?

(Isn't this first part an old puzzle?)

105^2=11025; 145^2=21025. In general we want 10^k=(y-x)(y+x) and

1.5 < (y/x)^2 < 2. Thus y+x and y-x must be factors of 10^k of

the same parity whose ratio is between 5.828... and 9.899...

(these are (t+1)/(t-1) for t^2=2 and 1.5 respectively). The

smallest solution (x,y)=(105,145) corresponds to the factorization

10^4=40*250; gp/pari's "fordiv" function allows one to easily list

all primitive solutions [i.e. not obtained from a smaller solution

by multiplying x,y by the same power of 10] with x^2 and y^2 each

having at most (say) 50 digits:

[x,y]=

[145, 105]

[17225, 14025]

[454625, 326625]

[53948125, 43708125]

[1425503125, 1015903125]

[168971890625, 136203890625]

[529265958203125, 424408358203125]

[1657888279384765625, 1322343959384765625]

[5193483785077392578125, 4119741961077392578125]

In fact it can be seen that the primitive solutions correspond to

integer linear combinations of log(2) and log(5) lying in a certain

fixed interval (determined by the bounds 5.828... and 9.899...),

which probably explains the regular growth of this list.

>2. What is the smallest square with leading digit 1 which remains a

>square when the leading 1 is replaced by a 2 and also remains a square

>when the leading digit is replaced by a 3?

There is no such beast, since the three squares would constitute an

arithmetic progression of integer squares of common difference 10^k,

and so give an A.P. of 3 rational squares of common difference 1 or 10 ---

which is known to be impossible by a "2-descent" argument (the case of

common difference 1 is already due to Fermat). [We were lucky here:

in a different number system this argument might fail; for instance the

squares of 7/2, 17/2, 23/2 are an A.P. of common difference 60, the

sexagesimal base. (Some numerology: 7,17,23 are the first three primes

of which 2 is a quadratic residue.) Still, given the base b, the general

theory of elliptic curves indicates that the rational solutions of

Y^2-X^2=Z^2-X^2=b are rather sparsely distributed (the number of d-digit

solutions growing as some power of d), and the extra condition that they

arise by changing only the initial digits of three integer squares is

strong enough to ensure that there are at most finitely many solutions;

with yet more powerful methods one can even provably list them all.]

>3. What is the smallest square with leading digit 1 which remains a

>square when the leading 1 is replaced by a 2, and also remains a square

>when the leading digit is replaced by a 3, and also remains a square

>when the leading digit is replaced by a 4?

Of course the above solution to part 2 also disposes of this part;

alternatively I could apeal to another classic result of Fermat:

there is no 4-term A.P. of rational squares.

My question: why all the blank spaces at the end of every line?

--Noam D. Elkies (elkies@zariski.harvard.edu)

Dept. of Math., Harvard Univ., Cambridge, MA 02138

-------------------------

I dunno the direct answer to your squares problem. I do know,

however, that phi (from the Golden Ratio--approx 0.61), which is

defined as the number x such that x + 1 = x^2. It just so happens

that phi+1 and (phi+1)^2 differ by only 1. (1.61 and 2.61) The

rest of the digits are the SAME! Phi = (Sqrt(5)-1)/2.

Phi+1 = (Sqrt(5)+1)/2 phi+2 = (Sqrt(5)+3)/2

(Phi+1)^2= (5+2*Sqrt(5)*1+1)/4 = (2*Sqrt(x)+6)/4 = (Sqrt(x) + 3)/2

Notice how that all works out? Perhaps this property will bring you

closer to an answer. I just sent you all my personal data in

a previous letter concerning your 123 problem. Let me know

what you think of this approach, ok? Thanks in advance!

--Joseph Zbiciak im14u2c@camelot.bradley.edu

-------------------------

In article <1992Oct20.184149.51596@watson.ibm.com> you write:

: 2. What is the smallest square with leading digit 1 which remains a

: square when the leading 1 is replaced by a 2 and also remains a square

: when the leading digit is replaced by a 3?

This is not possible. One of these numbers would leave a remainder

of 2 when divided by 3, and no square is congruent to 2 modulo 3.

--

David Radcliffe

radcliff@csd4.csd.uwm.edu

-------------------------

In article <1992Oct20.184149.51596@watson.ibm.com> you write:

: 1. What is the smallest square with leading digit 1 which remains a

: square when the leading 1 is replaced by a 2?

11025. I found, by hand, all integral solutions to

(x+y)(x-y) = 10000. The solution (145,105) is the only

one with 10000 < y^2 < 20000.

You have permission to use my solution, but not my name.

--

David Radcliffe

radcliff@csd4.csd.uwm.edu

-------------------------

Well, as a previous poster already mentioned on Rec.puzzles, there are only 4

solutions to the initial problem. They are 192, 219, 293, and 327. None of

these solutions can be connected to others as in part 2 of your problem.

I first extended the problem to allow any multipliers. So the second row must

be some multiple of the first and the third some other multiple of the first.

I found 19 solutions to this problem. However, there is still no way to chain

a second solution to the first.

Then I allowed 0s. Now there are 134 solutions. There are also 17 2-chains.

There are two 3-chains which I will list here:

192 394 *2= 384 *3=1182 *3= 576 *4=1576 *7=4032 now the same as the other solution. *9=5184 *4= 736 *5= 920

I will be more than happy to send you all 134 solutions if you really want

them! I also have Pascal source code.

Comments on some of your other problems will follow.

Dan Cory

Senior, Stanford

perm. address:

55 Cedar St.

Chapel Hill, NC 27514

school address:

PO Box 13113

Stanford, CA 94309

Should you use any of my results, please send a copy of the work to the

permanent address above.

-------------------------

In article <1992Oct20.184149.51596@watson.ibm.com>, you write:

|> Title: Cliff Puzzle 8: Squares and Squares and Squares ....

|> 1. What is the smallest square with leading digit 1 which remains a

|> square when the leading 1 is replaced by a 2?

11025 = 105^2, 21025 = 145^2.

|> 2. What is the smallest square with leading digit 1 which remains a

|> square when the leading 1 is replaced by a 2 and also remains a square

|> when the leading digit is replaced by a 3?

|>

|> 3. What is the smallest square with leading digit 1 which remains a

|> square when the leading 1 is replaced by a 2, and also remains a square

|> when the leading digit is replaced by a 3, and also remains a square

|> when the leading digit is replaced by a 4?

These two cases never occur.

Proof: (This was a LOT harder than I thought it would be when I started!)

The original problem can be reduced to:

"Find positive integers x,y,n such that

y^2-x^2 = 10^n and 10^n < x^2 < 2*10^n." [1]

The second problem amounts to finding x,y,z,n which meet the above

conditions, plus z^2-y^2=10^n.

For the second problem, look at the set of solutions to

z^2-y^2 = 10^n, 2*10^n < y^2 < 3*10^n. [2]

A solution to the second problem consists of x,y,z,n, where x,y,n solve

the original problem and y,z,n solve the above system.

The first equation in [1] can be factored into (y-x)(y+x) = 10^n = 2^n * 5^n.

Similarly (z-y)(z+y) = 10^n. Since x,y,z are integers, we must have

y+x = 2^a * 5^b, y-x = 2^(n-a) * 5^(n-b)

z+y = 2^c * 5^d, z-y = 2^(n-c) * 5^(n-d)

where a,b,c,d are integers. When a=c and b=d, y+x = z+y and y-x = z-y,

which leads to a contradiction.

Then 2y = 2^a * 5^b + 2^(n-a) * 5^(n-b) = 2^c * 5^d + 2^(n-c) * 5^(n-d)

However, in the last equality above, divide both sides by 2^f, where f is

the smallest of a, c, n-a, and n-c. The result is:

2^(a-f) * 5^b + 2^(n-a-f) * 5^(n-b) = 2^(c-f) * 5^d + 2^(n-c-f) * 5^(n-d) [3]

Now, at least one of the four products above is a product of only 5's, and is odd. Only one is odd unless a=c, 2a=n, or 2c=n. If a=c, then either b=d (contradiction) or z+y is at least a factor of 5 larger than y+x. However, considering sqrt(3)*sqrt(10^n) < z < 2*sqrt(10^n) sqrt(2)*sqrt(10^n) < y < sqrt(3)*sqrt(10^n) sqrt(10^n) < x < sqrt(2)*sqrt(10^n) we have: (sqrt(3)+sqrt(2))*sqrt(10^n) < z+y < (2+sqrt(3))*sqrt(10^n) (1+sqrt(2))*sqrt(10^n) < y+x < (sqrt(3)+sqrt(2))*sqrt(10^n) and then (z+y)/(y+x) < (2+sqrt(3))/(1+sqrt(2)) < 5. If a exactly equals n/2: In the case that b=a=n/2, y+x = y-x, so x=0 (not possible). If b<n/2, y-x>y+x, but we want x to be positive, so b>n/2. Since b and n/2 are integers (remember n/2=a), b-(n-b) >= 2, and (y+x)/(y-x) >= 25. This gives (y+x) >= 25(y-x), (y+x+y-x) = 2y >= 26(y-x), y >= 13y-13x, 13x >= 12y, x/y >= 12/13 x^2/y^2 >= 144/169

However, we know 10^n < x^2 < 2*10^n, and y^2 = x^2 + 10^n, so x^2/y^2

varies between 1/2 and 2/3, and cannot be greater than 144/169.

Similarly, when c=n/2, the same argument applies, and in the final step

we know y^2/z^2 varies between 2/3 and 3/4.

Finally, we've eliminated all cases where more than one of the terms in [3]

is odd. With exactly one term odd, we have odd=even, a contradiction,

so there is no solution.

-- ----w-w--------------Joseph De Vincentis--jwd2@owlnet.rice.edu---------------- ( ^ ) Disclaimer: My opinions do not represent those of Owlnet. (O O) Owlnet: George R. Brown School of Engineering Educational Network. v-v (Unauthorized use is prohibited.) (Being uwop-ap!sdn is allowed.) -------------------------

G'day Cliff!

> * * *

>

> 1. What is the smallest square with leading digit 1 which remains a

> square when the leading 1 is replaced by a 2?

>

> In other words, if x**2 = 1.........., is there a y**2 = 2......... ?

The smallest I could find was 105**2 = 11025

145**2 = 21025

Indeed, an exhaustive search shows that this is the smallest.

The other pairs I found (after a few minutes playing with pen and paper - I

could probably write a program to generate them ad nauseum, but I've got a

draft thesis to write...) were:

3375**2 = 11390625, 4625**2 = 21390625 14025**2 = 196700625, 17225**2 = 296700625 326625**2 = 106683890625, 454625**2 = 206683890625

I don't know what pattern there is in them. Of course, if x is a solution,

then so is 10*x. So these give solutions for 1050*1050 = 1102500, etc.

> 2. What is the smallest square with leading digit 1 which remains a

> square when the leading 1 is replaced by a 2 and also remains a square

> when the leading digit is replaced by a 3?

>

> 3. What is the smallest square with leading digit 1 which remains a

> square when the leading 1 is replaced by a 2, and also remains a square

> when the leading digit is replaced by a 3, and also remains a square

> when the leading digit is replaced by a 4?

I'll answer part 3 first. If such a square exists, then observe that we have

4 squares in arithmetic progression (common difference a power of 10). There

is a well known theorem that there is no set of four squares in arithmetic

progression, so there is no solution to part 3.

Now, for part 2. We have 3 squares in arithmetic progression. Another well

known (and not too hard to derive) theorem states that for three squares in

arithmetic progression, their common difference is of the form:

D = 4 * K^2 * m * n * (m^2 - n^2) = 4 * K^2 * m * n * (m + n) * (m - n)

Now, this value is a power of 10. So the only primes in its factorisation are

2 and 5. Hence neither m nor n is divisible by 3. So (m^2 - n^2) is

divisible by 3. Hence a power of 10 is divisible by 3. Contradiction. So

now such set of three squares exist (which also proves part 3).

Cheers,

Geoff.

PS: I assume you still have whatever details of mine you care about.

------------------------------------------------------------------------------- Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- ftww@cs.su.oz.au | Gameplayer by vocation. -------------------------------------------------------------------------------

-------------------------

Here is the solution I just posted to rec.puzzles. Note that I changed my mind

on this puzzle!

Dan Cory

Senior, Stanford

PO Box 13113

Stanford, CA 94309

ypay@leland.stanford.edu

Newsgroups: rec.puzzles

Subject: Re: Cliff Puzzle 8: Squares and Squares ... (SPOILER)

Approved: news-answers-request@MIT.Edu

Summary: solutions to part 1, no solutions to parts 2 or 3

Expires:

Sender:

Followup-To:

Distribution:

Keywords: squares, cliff, 8, gcd

In article <1992Oct20.184149.51596@watson.ibm.com> cliff@watson.ibm.com (cliff)

>1. What is the smallest square with leading digit 1 which remains a

>square when the leading 1 is replaced by a 2?

>In other words, if x**2 = 1.........., is there a y**2 = 2......... ?

We write this condition as the following equations with x,y,a integers:

y^2-x^2=10^a

1*10^a<=x^2<=2*10^a

2*10^a<=y^2<=3*10^a

We factor the first equation:

(y-x)(y+x)=10^a.

Let u=x+y. Then 10^a/u=x-y. Since x+y>x-y, u>10^a/u so u>10^(a/2)

Then x=(u-10^a/u)/2 and y=(u+10^a/u)/2.

Subsitute these equations into the inequalities above.

For x we get:

10^a<=((u-10^a/u)/2)^2<=2*10^a

Take the square root of both (all three?) sides:

10^(a/2)<=(u-10^a/u)/2<=sqrt(2)*10^(a/2)

Multiply through by 2 and divide through by 10^(a/2).

2<=u/10^(a/2)-10^(a/2)/u<=2*sqrt(2)

Let v=u/10^(a/2). So v>1. Then:

2<=v-1/v<=2*sqrt(2).

We solve these two inequalities. First the left:

v-1/v>=2

v^2-2v-1>=0

v>=(1+sqrt(2)) or v<=(1-sqrt(2)).

v-1/v<=2*sqrt(2)

v>=(sqrt(2)+sqrt(3)) or v<=(sqrt(2)-sqrt(3)).

Since v>1, we drop the negative solutions and find:

1+sqrt(2) <= v <= sqrt(2)+sqrt(3).

or

1+sqrt(2) <= u/10^(a/2) <= sqrt(2)+sqrt(3).

We can do the same for y but we will find the same restriction on u.

Now we remember that u|10^a (u divides 10^a). Therefore u must be a power of

2 times a power of 5. Let u=5^b*2^c with b,c integers less than or equal to a.

Since we are going to divide it by 2, we must have c>=1.

Then we need to find a,b,c such that:

1+sqrt(2) <= 5^b*2^c/10^(a/2) <= sqrt(2)+sqrt(3)

These will give us u which will in turn determine x and y.

So take the log base 10 of all three sides. Since log is increasing, we do not

change the direction of inequality. Thus:

log(1+sqrt(2)) <= b*log(5)+c*log(2)-a/2 <= log(sqrt(2)+sqrt(3))

Multiply through by 2:

2*log(1+sqrt(2)) <= 2*b*log(5)+2*c*log(2)-a <= 2*log(sqrt(2)+sqrt(3))

If we approximate log(5) and log(2), this is sort of a Diophantine equation.

Since log(5) is very very close to 0.7 and log(2) is very very close to 0.3,

our approximations will be okay to find low solutions. If we want big solutions

then we need to use better convergents. We can calculate the boundary logs

as accurately as necessary. So:

0.77 <= 7/5*b+3/5*c-a <= 0.99

Multiply through by 5:

3.8 <= 7*b+3*c-5*a <= 4.9

So we must find a,b,c such that 7*b+3*c-5*a = 4, with a>b>=0 and a>=c>0.

There are many good ways to solve this but we will just pick a small solution.

b=3, c=1, a=4 (7*3+3-5*4=21+3-20=4)

Then u=5^3*2^1=250.

So y+x=250 and y-x=10^a/u=10^4/250=40.

Then y=145 and x=105.

y^2=21025 and x^2=11025.

This is, in fact, the smallest solution (it is easy to show that there is no

solution to the 7*b+3*c-5*a with a<4 and a>b>=0,a>=c>0).

>2. What is the smallest square with leading digit 1 which remains a

>square when the leading 1 is replaced by a 2 and also remains a square

>when the leading digit is replaced by a 3?

We note from above that y=(5^b*2^c+10^a/(5^b*2^c)/2 or

2y=5^b*2^c+5^(a-b)*2^(a-c).

Should we now repeat the problem for a square with leading digit 2 that is

replaced by a 3, everything is the same except that y is now the smaller of the

pair. Thus:

2y=5^B*2^C-5^(a-B)*2^(a-C)

where B and C are different from b and c above but a is necessarily the same

(since we want the difference to be the same power of 10 for each transition).

Combining the two we get:

5^b*5^c+5^(a-b)*2^(a-b)=5^B*2^C-5^(a-B)*2^(a-C).

The proof that this has no solutions is too small to fit in the margin of this

posting.

>3. What is the smallest square with leading digit 1 which remains a

>square when the leading 1 is replaced by a 2, and also remains a square

>when the leading digit is replaced by a 3, and also remains a square

>when the leading digit is replaced by a 4?

There is no solution since there is no solution to part 2.

Dan Cory

Continue to: