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,…

For 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…

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

Two smallest such numbers are

10894620496601400001

24999000019999800001