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A261782
Powers C^z = A^x + B^y with positive integers A,B,C,x,y,z such that x,y,z > 2.
4
16, 32, 64, 128, 243, 256, 512, 1024, 2048, 2744, 4096, 6561, 8192, 16384, 32768, 65536, 131072, 177147, 185193, 262144, 474552, 524288, 614656, 810000, 941192, 1048576, 1124864, 1419857, 1500625, 2097152, 3241792, 4194304
OFFSET
1,1
COMMENTS
Beal's conjecture states that A, B, and C have a common prime factor.
Theorem. If A, B are odd and x, y are even, Beal's conjecture has no counterexample. Proof: Let D be odd, D > 1 and let w be even, w > 2. Then D^w == 9 (mod 24) while D == 0 (mod 3); otherwise, D^w == 1 (mod 24) (trivial). Any even C^z == {0; 8; 16} (mod 24): if C == 0 (mod 3), C^z == 0 (mod 24); if C == 1 (mod 3), C^z == 16 (mod 24); if C == 2 (mod 3), C^z == 8 (mod 24), while z is odd, and C^z == 16 (mod 24), while z is even (trivial). But C^z == (x'+y') (mod 24) where A^x = x' (mod 24), B^y = y' (mod 24); since (x'+y') = {2; 10; 18}, C^z == {2; 10; 18} (mod 24), which cannot be a counterexample to Beal's conjecture. - Sergey Pavlov, May 08 2021
LINKS
Anatoly E. Voevudko and Charles R Greathouse IV, Table of n, a(n) for n = 1..1229 (first 196 terms from Voevudko)
American Mathematical Society, Beal Prize
EXAMPLE
2^3 + 2^3 = 2^4 = 16, so 16 is in the sequence.
PROG
(PARI) is(n)=if(ispower(n)<3, return(0)); for(x=3, logint((n+1)\2, 2), for(A=2, sqrtnint(n, x), if(ispower(n-A^x)>2, return(1)))); 0 \\ Charles R Greathouse IV, Sep 03 2015
(PARI) list(lim)=my(v=List(), u=v, t); for(z=3, logint(lim\=1, 2), for(C=2, sqrtnint(lim, z), listput(v, C^z))); v=Set(v); for(i=1, #v, for(j=i, #v, t=v[i]+v[j]; if(t>lim, break); if(setsearch(v, t), listput(u, t)))); Set(u) \\ Charles R Greathouse IV, Sep 03 2015
CROSSREFS
Subsequence of A076467.
Cf. A245713.
Sequence in context: A317475 A335161 A239751 * A256818 A048170 A340624
KEYWORD
nonn
AUTHOR
Anatoly E. Voevudko, Aug 31 2015
STATUS
approved