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A200519
Least m>0 such that n = y^2 - 9^x (mod m) has no solution, or 0 if no such m exists.
1
0, 4, 4, 0, 8, 4, 4, 0, 0, 4, 4, 9, 8, 4, 4, 0, 0, 4, 4, 0, 8, 4, 4, 9, 0, 4, 4, 0, 8, 4, 4, 80, 9, 4, 4, 0, 8, 4, 4, 63, 0, 4, 4, 80, 8, 4, 4, 9, 0, 4, 4, 45, 8, 4, 4, 0, 9, 4, 4, 9, 8, 4, 4, 0, 133, 4, 4, 80, 8, 4, 4, 15, 0, 4, 4, 63, 8, 4, 4
OFFSET
0,2
COMMENTS
To prove that an integer n is in A051211, it is sufficient to find integers x,y such that y^2 - 9^x = n. In that case, a(n)=0. To prove that n is *not* in A051211, it is sufficient to find a modulus m for which the (finite) set of all possible values of 9^x and y^2 (mod m) allows us to deduce that y^2 - 9^x can never equal n. The present sequence lists the smallest such m>0, if it exists.
EXAMPLE
See A200512 for motivation and detailed examples.
PROG
(PARI) A200519(n, b=9, p=3)={ my( x=0, qr, bx, seen ); for( m=3, 9e9, while( x^p < m, issquare(b^x+n) & return(0); x++); qr=vecsort(vector(m, y, y^2-n)%m, , 8); seen=0; bx=1; until( bittest(seen+=1<<bx, bx=bx*b%m), for(i=1, #qr, qr[i]<bx & next; qr[i]>bx & break; next(3))); return(m))}
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Nov 18 2011
STATUS
approved