I looked at Euler Problem 808 (https://projecteuler.net/problem=808) and made a strange observation (please don't post help solving the problem):
For an integer x, let r(x) be the number created by writing the decimal digits of x in reverse order. For example r(123456) = 654321, r(1100220033) = 3300220011. The problem requires to find primes n such that r(n^2) is the square of a different prime m. There are about 130 solutions for n < 3.1 * 10^9 in 65 pairs. The first pair is 13^2 = 169 and 31^2 = 961.
The strange observation: For all pairs except 3 I found that m = r(n). The only exceptions are
110092211^2 = 12120294922868521, 112191011^2 = 12586822949202121
1100091121^2 = 1210200474503036641, 1210910011^2 = 1466303054740020121
1100910121^2 = 1212003094520234641, 1210091011^2 = 1464320254903002121
Is that coincidence, or can anyone see a reason why these numbers are so rare (numbers where m^2 = r(n^2) but m != r(n))? If I change the requirement from "n, m are prime" to "n, m end in 1, 3, 7 or 9", then I get about 2115 pairs of squares with 127 pairs where m != r(n), a very similar percentage.
Note: Generalizing $\,r(n^2) = r(n)^2,\,$ for $\,r(n) := $ reverse the digits of $n$ shows that if you interpret a decimal number with k digits as a polynomial of degree k-1, evaluated at x = 10, then squaring n and reversing the result, and reversing n and squaring the result, will always give the same polynomial. If all the coefficients are less than 10, then we also have the same number. (But that is only part of the question).
So the explanation may be that while having a polynomial with all coefficients less than 10 is rare, having numbers where the digits are all right by coincidence is even less rare. I'll look at that and try to post an answer tonight.
