THE NUMBER OF REPRESENTATIONS OF n OF THE FORM n = x 2 2 y, x > 0, y 0

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1 THE NUMBER OF REPRESENTATIONS OF n OF THE FORM n = x 2 2 y, x > 0, y 0 RICHARD J. MATHAR Abstract. We count solutions to the Ramanujan-Nagell equation 2 y +n = x 2 for fixed positive n. The computational strategy is to count the solutions separately for even and odd exponents y, and to handle the odd case with modular residues for most cases of n. 1. Scope We consider the number of ways to represent n > 0 as (1) n = x 2 2 y, for pairs of the non-negative unknowns x and y. Equivalently we count the y such that (2) n + 2 y = x 2 are perfect squares. Remark 1. For n = 0 the number of solutions is infinite because each even value of y creates one solution. The n for which the count is nonzero end up in [8, A051204]. Remark 2. Negative n for which solutions exists are listed in [8, A051213] [4, 2]. The solutions are counted separately for even y and odd y and added up. The total count is at most 4 [6]. 2. Even Exponents y Counting the squares x 2 of the format (2) considering only even y is basically a matter of considering the values y = 0, 2, 4,... in turn and checking explicitly each y 2 + n against being a square. An upper limit to the y is determined as follows: The y-values in the range 2 y < n are all checked individually. For larger y the values of 2 y + n on the left hand side of (2) are represented in binary by some most significant bit contributed by 2 y and after a train of zero bits depending of how much larger 2 y is than n the trailing bits of n. A lower exact bound of the x on the right hand side is x = 2 y/2, contributing to the count if n were zero. So the next higher candidate on the right hand side is (x+1) 2 = (2 y/2 +1) 2 = 2 y +2 1+y/2 +1. If this value is larger than the left hand side 2 y + n, there are no further solutions because Date: October 19, Key words and phrases. Ramanujan-Nagell, Exponential Diophantine. 1

2 2 RICHARD J. MATHAR n is so small that it falls into the gap between consecutive squares. In summary, the search range for this branch of the algorithm can be reduced to 2 1+y/2 + 1 n. The sequence of the number of representations of n 1 with even y (i.e., the number of representations n = x 2 4 y ) is 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1,.... The smallest n which has r representations of the form n = x 2 4 y is the greedy inverse of this sequence, which is n = 1 for r = 0, n = 3 for r = 1, n = 33 for r = 2, n = 105 for r = 3 and n = 1680 for r = 4. There are no n for r 5 [6]. 3. Odd Exponents y The count of solutions to (2) with odd y is split into cases (i) for n being multiples of 4, (ii) n represented by at least one modulus m such that an upper limit of y is found by considering the equation modulo m, and (iii) other (discriminants) n that apparently do not fall into these categories n which are multiples of 4. If 4 n in (2) for odd y = 2y o + 1, the left hand side is even, so the right hand side and x = 2x e must also be even. Then considering the equation modulo 4, the case y o = 0 cannot yield a solution, and one may divide each term of the equation by 4. This reduces recursively the number of solutions to the number of solutions for n/ n with associated moduli. The sequences 2 y + n = 2 2ye+1 + n (y e = 0, 1, 2,...) and x 2 (x = 0, 1, 2,...) on both sides of (2) have generating functions which are rational functions and therefore have Pisano periods if read reduced some modulus m [7, 10, 11]. Remark 3. For the sequence x 2 [8, A000290] the length of the Pisano periods is tabulated in [8, A186646] as a function of m. For polynomial sequences like x 2, the period length is obviously limited to m and always a divisor of m. Discarding duplicates in the period, the number of squares modulo m is also finite [8, A000224]. Remark 4. If 2 and m are coprime, the length of the Pisano period of 2 y is the multiplicative order of 2 (mod m) [8, A002326]. The sequences 2 y + n (mod m) generally start with a transient list of moduli at small y before entering the period. Example 1. The sequence 2 y (mod 68) reads 1, 2, 4, 8, 16, 32, 64, 60, 52, 36, 4, 8, 16, 32, 64, 60, 52,... for y 0 with a transient part containing 1 and 2 and a periodic part containing 4, 8, 16, 32, 64, 60, 52, 36. The computational strategy is to find a modulus m given n such that the two periods of 2 2ye+1 + n (mod m) and of x 2 (mod m) have no common element. If such a modulus m is found, the length of the transient part of the 2 y + n residua defines the maximum exponent y that needs to be searched, because for all larger y the moduli of both sides of (2) are distinct. Remark 5. There is no transient part in the residua of the polynomials like x 2.

3 THE NUMBER OF REPRESENTATIONS OF n OF THE FORM n = x 2 2 y, x > 0, y 0 3 Cases of a search for m for some small n are illustrated in Table 1. Only successful m are listed, as indicated by distinct sets of moduli in the penultimate and ultimate column in the table. Cases where n is a multiple of 4 are left out because the reduction of Section 3.1 makes them uninteresting. Remark 6. Chosing m = 3 works if 3 n because 2 y (mod 3) = 1, 2,... (y 0) and because x 2 (mod 3) = 0, 1, 1,..., (x 0) both periodically repeated. Since we are considering only odd y, the set of residua contains {2} for 2 y and {0, 1} for x 2, and these do not intersect. The cases of odd n which are not in the table are apparently not accessible by the method of residue and are discussed in Section 3.3. Note that a shortcut exists for n which are two times an odd number, n = 2n o. Then 2 y + n = 2 2yo+1 + 2n o = x 2 requires x to be even, x = 2x e, so 2 2yo+1 + 2n o = 4x 2 e and 2 2yo + n o = 2x 2 e. Since the right hand side is even, the left hand side is even which requires 2 2yo to be odd and y o = 0 or no solution at all. This is decided by direct inspection. (See [8, A056220] for the n o that do have a solution.) 3.3. Other n. For n = 1, 17, 41, 49, 73, 89,... no such modulus m has been found that separates the residue sets of 2 y + n and x 2. These n are discussed individually [9]: For n = 1 only the solution with y = x = 3 exists [1]. For n = 17 the maximum number of 4 solutions [5, 9] is known, represented by (x, y) = (5, 3), (7, 5), (9, 6) and (23, 9). (One of these is created by an even y and already counted in Section 2.) For n = 41 we have (x, y) = (7, 3) or (13, 7). Because 41 is in [8, A031396], the associated equation u 2 41v 2 = 1 has solutions (explicit u = 32 in [8, A249021]). and according to Le s second theorem [5] there are no more than 2 solutions. For n = 49 we have (x, y) = (9, 5). This is the only solution. [Proof: Solving 2 y + 49 = x 2 for y 1 needs odd x by considering the parity of both sides. So this is 2 y = (x + 7)(x 7) where x ± 7 are both even. Furthermore comparison of the prime factorization of both sides enforces that x ± 7 are powers of 2, say x 7 = 2 α, x + 7 = 2 α+δ. Subtraction of both equations gives 14 = 2 α (2 δ 1) = 2 7. Necessarily α = 1, β = 3 and finally x = 9. See [3].] For n = 73 we have (x, y) = (9, 3). This is the only solution [9] For n = 89 we have (x, y) = (11, 5) or (91, 13). Again 89 is in [8, A031396] and these are all solutions according to Le s second theorem [5]. For n = 97 we have (x, y) = (15, 7) as the only solution [9]. For n = 113 we have (x, y) = (11, 3) or (25, 9). Again 113 is in [8, A031396] and these are all solutions according to Le s second theorem [5]. For n = 161 the maximum number of 4 solutions [5] is known, represented by (x, y) = (13, 3), (15, 6), (17, 7) and (47, 11). (One of these is created by an even y and already counted in Section 2.) For n = 833 the maximum number of 4 solutions [5] is known, represented by (x, y) = (29, 3), (31, 7), (33, 8) and (95, 13). (One of these is created by an even y and already counted in Section 2.)

4 4 RICHARD J. MATHAR Table 1. Examples of the dividing property of moduli m for small n, odd y. The double column entitled 2 y + n shows the transient values of 2 y + n (mod m) and the values of 2 y + n (mod m) in the period. The column entitled x 2 shows the period of x 2 (mod m) (of length m, not necessarily reduced to the smallest subperiod). n m π(2 y + n) π(x 2 )

5 THE NUMBER OF REPRESENTATIONS OF n OF THE FORM n = x 2 2 y, x > 0, y 0 5 References 1. K. Chao, On the diophantine equation x 2 = y n + 1, xy 0, Sci. Sinica (Notes) 14 (1964), J. H. E. Cohn, The diophantine equation x 2 +c = y n, Acta Arithm. 65 (1993), no. 4, MR , The diophantine equation x k = y n, ii, Int. J. Math. Math. Sci. 22 (1999), no. 3, Hemar Godinho, Diego Marques, and Alain Togbé, On the diophantine equation x 2 = 2 α 5 β 17 γ = y n, Commun. Math. 20 (2012), no. 2, MR Maohua Le, On the number of solutions of the generalized Ramanujan-Nagell equation x 2 d = 2 n+2, Acta Arithm. 60 (1991), no. 2, MR , A note on the number of solutions of the generalized Ramanujan-Nagell equation x 2 d = k n, Acta Arithm. 78 (1996), no. 1, MR D. W. Robinson, A note on linear reccurent sequences modulo m, Am. Math. Monthly 73 (1966), no. 6, MR Neil J. A. Sloane, The On-Line Encyclopedia Of Integer Sequences, Notices Am. Math. Soc. 50 (2003), no. 8, , MR (2004f:11151) 9. Nicholas Tzanakis, On the diophantine equation y 2 d = 2 k, J. Number Theory 17 (1983), no. 2, Morgan Ward, The characteristic number of a sequence of integers satisfying a linear recursion relation, Trans. Am. Math. Soc. 33 (1931), no. 1, MR , The arithmetical theory of linear recurring series, Trans. Am. Math. Soc. 35 (1933), no. 3, MR URL: Max-Planck Institute of Astronomy, Königstuhl 17, Heidelberg, Germany

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