[ October 14, 2021 ]
Playing around with recreational examples of various types
like combinations using squares, factorials and other functions
Vinod Muleva (email)

Ex. (4!)² + (5!)² = 101² + 69² + 3² + 2² + 1²
5! * 6! * 7! = 757³ + 118³ + 24³ + 10³ + 3³ + 3 * 2³
6! – 5! + 4! – 3! + 2! – 1! = 619 (Prime!)
6! + 5! – 4! + 3! – 2! – 1! = 821 (Prime!)
10! – 9! – 8! – 7! – 6! – 5! – 4! – 3! – 2! – 1! = 1794² – 35² – 5² – 1²

Vinod gave also some random examples of palindromes
that are the sum of powers.
12321² + 121² – 117² + 11² + 6² + 1² = 151808151
10001² + 11311² + 10⁴ + 10³ = 227969722

In an effort to structure these terms I propose the following challenge.

Find for all the squares s1 and the 'next' higher square s2 (s2 > s1) so
that their sum is palindromic.

Let me start with some initial terms of s1 + s2 = Pal
1² + 2² = 5
2² + 20² = 404
3² + 18² = 333
4² + 14² = 212
5² + ? = ?
6² + ? = ?
7² + ? = ?
...
Maybe some squares will have an accompanying square s2 far far greater than s1 !!
I wonder if it stays easy to find these s2 squares.

Hope you like my challenge. If so can you do this also for third powers ?

In the mean time enjoy also these equations from Vinod
turning the palindromes of the form 10^n–1 into a sum of powers

I guess these are the minimum number of terms (squares)
with which to express the palindromes ?
If someone can do it with even less terms (powers besides
squares allowed), please let me know.

[ December 12, 2021 ]
The ink was not yet dry or Alexandru Petrescu
improved these Vinod equations — with only three terms.
Well done!