A051204 -id:A051204 - cp's OEIS frontend

A200519 Least m>0 such that n = y^2 - 9^x (mod m) has no solution, or 0 if no such m exists.

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

M. F. Hasler, Nov 18 2011

nonn

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.

			See A200512 for motivation and detailed examples.

Cf. A051204-A051221, A200505-A200520.

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 & break; next(3))); return(m))}

A362564 a(n) is the largest integer x such that n + 2^x is a square, or -1 if no such number exists.

Original entry on oeis.org

3, 1, 0, 5, 2, -1, 1, 3, 4, -1, -1, 2, -1, 1, 0, 7, 9, -1, -1, 4, 2, -1, 1, 0, -1, -1, -1, 3, -1, -1, -1, 5, 8, 1, 0, 6, -1, -1, -1, -1, 7, -1, -1, -1, 2, -1, 1, 4, 5, -1, -1, -1, -1, -1, -1, 3, 6, -1, -1, 2, -1, 1, 0, 9, 10, -1, -1, 11, -1, -1, -1, -1, 3, -1, -1, -1, 2, -1, 1, 6, -1, -1, -1, 4, -1

Offset: 1

Yifan Xie, Apr 24 2023

sign

a(n) is the maximum integer solution x of the indefinite equation n + 2^x = r^2 where n is a constant and r is a positive integer, or -1 if there are no solutions.
See A247763 for the number of solutions and for further information, references and links about this problem.
If n == 0 (mod 4), we first try x = 0 or 1; when x >= 2, the result can be derived from the result for n/4 (see formula).
If n == 2 (mod 4), the only possible values of x is 1, as otherwise n + 2^x == 2 (mod 4), so nonsquare.
If n == 3 (mod 4), the only possible values of x are 0 and 1, as otherwise n + 2^x == 3 (mod 4), so nonsquare.
If n == 1 (mod 4), or n = 4*k + 1 (k >= 0): we suggest that r = 2*m + 1, 4*k + 1 + 2^x = (2*m + 1)^2, thus k + 2^(x - 2) = m*(m + 1); if k is odd, the only possible values of x is 2 because m*(m + 1) is even.
If n = k^2 (k >= 1), 2^x = (r + k)*(r - k), so 2k must be in the form 2^i - 2^j.
The problem can be solved via reduction to three Mordell curves: n + z^3 = y^2, n + 2z^3 = y^2 or equivalently 4n + (2z)^3 = (2y)^2, n + 4z^3 = y^2 or equivalently 16n + (4z)^3 = (4y)^2, where z := 2^floor(x/3). For a given n, each of these three curves is known to have only a finite number of integer points (y,z), proving that x cannot be unbounded. - Max Alekseyev, Apr 26 2023

			See COMMENTS section for further proof.
For n = 1, 1 + 2^3 = 9 = 3^2;
for n = 4, 4 + 2^5 = 36 = 6^2;
for n = 7, 7 + 2^1 = 9 = 3^2;
for n = 9, 9 + 2^4 = 25 = 5^2.

Thomas Scheuerle, Table of n, a(n) for n = 1..1000

Cf. A000079, A000290, A051204, A234000, A238454, A247763.

def a362564(n): return max((3*v-2*k for k in range(3) for z,, in EllipticCurve([0,4^k*n]).integral_points() if z>=1<Max Alekseyev, Apr 26 2023

a(4*k + 2) = 1 if k + 1 is a square, or -1 otherwise.
a(4*k + 3) = 1 if 4*k + 5 is a square, or 0 is k + 1 is a square and 4*k + 5 is a nonsquare, or -1 otherwise.
a(4*k + 4) = a(k) + 2 if a(k) >= 0, or 0 if 4*k + 1 is a square and a(k) = -1, or -1 otherwise.
a(8*k + 5) = 2 if 8*k + 9 is a square, or -1 otherwise.
a((2^i - 2^j)^2) = i + j + 2 for i,j >= 0.
a(n) > -1 if A247763(n) > 0, or equivalently n is in A051204. - Thomas Scheuerle, May 02 2023

cp's OEIS Frontend

A200519 Least m>0 such that n = y^2 - 9^x (mod m) has no solution, or 0 if no such m exists.

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