OFFSET
1,1
COMMENTS
Solutions to 19^x + 23^x == 29 mod 31.
The form of the these numbers is obviously 30X + 13. 3^x + 5^x == 7 mod 11 and 17^x + 19^x == 23 mod 29 have no solutions. In fact, 3^x + 5^x == m mod 11 is only solvable for m < 11 = 1, 2, 8, 9. Similarly, 17^x + 19^x == m mod 29 is not solvable for m < 29 = 6, 11, 13, 14, 15, 16, 18, 23. I can't even prove 3^x + 5^x-7 <> 11k for all integers x, k. Anyone have a general proof of these statements say, a^x + -b^x == m mod k true or false for certain a, b, m, k, x combinations?
a^x + b^x == m (mod k) is periodic mod phi(k), so it suffices to check x = 1, 2, ..., phi(k). - Charles R Greathouse IV, Nov 19 2013
LINKS
Michael G. Kaarhus, Table of n, a(n) for n = 1..10000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2,-1).
FORMULA
a(n) = 30n + 13.
G.f.: x*(13+17*x)/(1-x)^2. - Colin Barker, Jan 11 2012
MATHEMATICA
Range[13, 7000, 30] (* Vladimir Joseph Stephan Orlovsky, Jul 13 2011 *)
LinearRecurrence[{2, -1}, {13, 43}, 50] (* Harvey P. Dale, Mar 02 2023 *)
PROG
(PARI) anpbn(n)= for(x=1, n, if((19^x+23^x-29)%31==0, print1(x" "))) \\ solutions to 19^x+23^x == 29 mod 31
(Maxima) f(a):= mod((19^a + 23^a), 31)$ a:-1$ for n:1 thru 3000 step 0 do(a:a+1, if f(a)=29 then (if mod(a, 30)=13 then (print(n, " ", a), n:n+1) else (print("Exception at ", a, ", ", f(a)), n:3001))) /* f(a)==29 only when a is cong. to 13 (mod 30). No exceptions thru a=89983, n=3000. Michael G. Kaarhus, Nov 18 2013 */$
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Cino Hilliard, May 11 2003
EXTENSIONS
Simpler name from Charles R Greathouse IV, Nov 19 2013
STATUS
approved