

A115038


Smallest x such that x^2 + p is a perfect power over the set of primes p.


0



5, 1, 2, 1, 4, 6, 8, 9, 2, 14, 1, 18, 20, 21, 9, 26, 29, 8, 33, 21, 36, 7, 41, 6, 48, 50, 5, 53, 4, 56, 1, 65, 68, 69, 74, 19, 78, 81, 7, 86, 8, 90, 5, 96, 98, 12, 105, 17, 4, 114, 116, 2, 120, 44, 128, 131, 134, 27, 138, 140, 141, 146, 6, 155, 156, 158, 165, 168, 173, 174, 176, 37
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OFFSET

1,1


COMMENTS

Conjecture: There will always be an x,y,n such that x^2 + p = y^n for all primes p. In otherwords, there is a oneone mapping of the prime numbers to y^n  x^2 for some x,y,n. Therefore primes of the form y^n  x^2 are infinite in number.


LINKS



EXAMPLE

5 is the smallest number that when we add its square to prime 2, we get a
perfect power, 3^3. 1 is the smallest number that when we add its square to
prime 3, we get a perfect power, 2^2. So 5 and 1 are the first two entries in
the table.


PROG

(PARI) sqplusp(n) = { local(p, x, y, c=0); forprime(p=2, n, for(x=1, n, y=x^2+p; if(ispower(y), print1(x", "); c++; break ) ) ); print(); print(c) }


CROSSREFS



KEYWORD

easy,nonn


AUTHOR



STATUS

approved



