## Four Puzzles for the Price of One

Here is a math problem from the 1977 USSR Math Olympiad:

Let *A* be a *2n*-digit number. We call this number **special** if it is a square and a concatenation of two *n*-digits squares. Also, the first *n*-digit square can’t start with zero; the second *n*-digit square can start with zero, but can’t be equal to zero.

- Find all two- and four-digit special numbers.
- Prove that there exists a 20-digit special number.
- Prove that not more than ten 100-digit special numbers exist.
- Prove that there exists a 30-digit special number.

Obviously, these questions are divided into two groups: show the existence and estimate the bound. Furthermore, this problem can be naturally divided into two other groups. Do you see them? The puzzle about special numbers makes a special day today — you get a four-in-one puzzle.

Share:
## Zak Seidov:

As to

* Prove that there exists a 20-digit special number.

Two smallest such numbers are

24 October 2008, 2:17 am10894620496601400001

24999000019999800001

## Zak Seidov:

no more 20-digit special numbers!

24 October 2008, 2:46 pm## Philippe Fondanaiche:

The numbers 5*10^k – 1 and 10^(k+1) – 1 provide (4k+4)-digit special numbers for k = 0,1,2,3,…

Indeed 10^(2k+2)*(5*10^k – 1)^2 + (10^(k+1) – 1)^2 = (5*10^(2k+1) – 10^(k+1) + 1)^2

So there always exist special numbers whose length is a multiple of 4, particularly length = 20 and 100.

Examples 4^2 + 9^2 = 41^2, 49^2 + 99^2 = 4901^2, 499^2 + 999^2 = 499001^2, etc….,the above mentioned solution 49999^2 + 99999^2 = 4999900001^2

On the other hand, solutions can be found on the basis of the Pell equations of the form 10*x^2 + p^2 = y^2 with p = 1, 2, 3 or 1000*x^2 + p^2 = y^2 with p = 11,12,…

14 May 2009, 5:19 pmFor example we can obtain: x=18, p = 3 and y = 57 with the corresponding special number 324 900 = 570^2

or x = 228, p = 1, y = 721 with the special number 5 198 410 000 = 72100^2

etc…