Computational algorithm for solving the diophantine equations 2n±α·2m+α2 =x2
L´aszl´o Szalay J. Selye University
94501 Kom´arno, Bratislavsk´a cesta 3322, Slovakia laszlo.szalay.sopron@gmail.com
Abstract
In this paper, we construct an algorithm for solving the diophantine equations 2n±α·2m+α2=x2, whereα is a given odd prime such that 2 is a non-quadratic residue modulo α. Applying the implementation of the procedure in Maple, apart from the plus case with the condition n < m we solve completely the problem for α <3·106. The theoretical background relies on the treatment worked out to solve the equation 2n+ 2m+ 1 =x2.
AMS Subject Classification: 11D61.
Keywords: algorithm, diophantine equations, polynomial-exponential equations, num- ber of bits in squares.
1 Introduction
There exist several works on studying the occurrence of perfect squares in specified infinite sets of integers having only few nonzero digits in basep. For example, the simple equation 2n+ 1 =x2 with positive integers n asks for the odd integers x having two 1 bits in the binary expansion of its square. The analogous equation with three 1 bits was studied in [8]. The author proved that the equation 2n+ 2m + 1 = x2 with integers n ≥ m ≥ 0 and x ≥ 0 has only the solutions (n, m, x) = (2t, t+ 1,2t + 1) for integers t ≥ 1, and (n, m, x) = (1,0,2), (5,4,7), (9,4,23). The approach of this paper is built upon the method worked out in [8]. The equation 2n−2m + 1 = x2 was also considered by [8].
The solutions are (n, m, x) = (2t, t+ 1,2t−1) (t ≥ 2), (n, m, x) = (t, t,1) (t ≥ 1), and (n, m, x) = (5,3,5), (7,3,11), (15,3,181). Luca [7] extended the problem to arbitrary odd prime base p, and proved that the equation pn +pm+ 1 = x2 possesses no integer solutions.
The question arises naturally: what happens if the square is replaced by any pure power? The paper [1] of Bennett and Bugeaud contains the following result. If sn + sm+ 1 = xk holds for the positive integers s, n > m, k ≥2 with gcd(k, ϕ(s)) = 1, then (s, n, m, xk) = (2,5,4,72), (2,9,4,232), (3,7,2,133), or (2,2t, t+ 1,(2t + 1)2) (t ≥ 1).
When the condition gcd(k, ϕ(s)) = 1 is omitted, but s= 2 or 3 is assumed, the equation has the same set of solutions (see [2] and [1]).
Another extension of the problem is to take different bases of powers. In this direction Hajdu and Pink [4] completely solved the diophantine equation 1 + 2a+tb =xk assuming odd t≤50. Later B´erczes et al. [5] examined the more general equation 1 +sa+tb =xk, and provided all the solutions with the conditionsk≥4, 1≤s, t ≤50, ands 6≡t (mod 2).
Now consider the following variation of the equations 2n±2m+ 1 =x2. Multiply the second 2-power by an odd prime α, and replace the constant 1 term by α2 to obtain the title equations. We assume hereα≥5 since the caseα= 3 was handled in [6]. Further we suppose that the Legendre symbol (2|α) =−1, i.e. 2 is a quadratic non-residual modulo α. Such primes have the form 8β±3, the first few are
3,5,11,13,19,29,37,43,53,59,61,67,83,101, . . . . The principal result of this paper is an algorithm for solving
2n±α·2m+α2 =x2 (1)
with given α, respectively, in the + case with the condition n ≥m. We implemented the procedures in Maple, and ran them for α < 3·106 in each case. Clearly, x = |2t±α|
always satisfies (1), we call this family regular solutions. Hence the algorithm determines the exceptional (i.e. non-regular) solutionsE±, respectively. The results are recorded here in two tables (Table 1, and Table 2), and later in Theorem 4, and 7.
α (n, m, x) α (n, m, x)
3 (6,3,7) 5 (4,0,6)
5 (6,3,7) 11 (8,3,17)
13 (10,3,33) 37 (14,4,131) 53 (14,3,137) 149 (18,4,531) 853 (22,3,2217) 2389 (26,4,8531) 3413 (26,3,8873) 57467 (42,4,2097939) 218453 (38,3,567977) 2708477 (66,4,8589935019)
Table 1: solution to 2n−α·2m+α2 =x2 with α∈P, (2|α) =−1,α <3·106 To close this section we recall a very useful lemma due to Beukers [3] (Corollary 2).
It gives an upper bound on n, therefore it is effective, and we will use it several times in the algorithms.
Lemma 1. Let 06=D∈Z. If |D|<296 and 2n+D=x2 has a solution (n, x), then n <18 + 2 log2|D|.
Note that Corollary 1 of [3] provides the general upper bound n <435 + 10 log2|D|,
which is valid for arbitrary non-zero integer D.
Along this paper,ν2(n) will denote the largest exponent ν of 2 such that 2ν |n.
α (n, m, x) α (n, m, x)
3 (2,0,4) 3 (6,5,13)
3 (8,3,17) 5 (4,4,11)
5 (8,4,19) 5 (10,3,33)
11 (6,0,14) 11 (10,0,34)
13 (8,3,23) 19 (12,4,69)
19 (16,3,257) 29 (14,4,133)
43 (10,0,54) 107 (16,3,279)
149 (24,4,4099) 317 (22,3,2073) 461 (22,4,2101) 683 (18,0,854) 2731 (22,0,3414) 6827 (28,3,17751) 43691 (30,0,54614) 44939 (34,0,138562) 174763 (34,0,218454) 1887437 (46,4,8598325) 2796203 (42,0,3495254)
Table 2: solution to 2n+α·2m+α2 =x2 with n ≥m, α∈P, (2|α) =−1, α <3·106
2 Preliminary considerations, τ transformation
Consider equations (1) moduloα. The Legendre symbol (2|α) = −1 yields thatn is even.
Put R± = {(n, m, x) ∈ N3 | n = 2t, m = t+ 1, x = |2t ±α|, t ≥ 0}, respectively.
Obviously, the elements of R± satisfy (1), respectively. They are called regular solutions.
Let A± denote the set of all solutions to (1), respectively. The main goal is to determine E±=A±\R±, the set of exceptional solutions, respectively.
Assume m ≥1. Thus x is odd. Furthermore, suppose n≥ m. Observe that if (n, m) is a solution to (1), then 2n−m±α= (x2−α2)/2m ∈N, moreover
22n−2m±α·2n−m+1+α2 =
x2−α2 2m
2
.
Hence a solution (n, m) of (1) provides the solution (2n−2m, n−m + 1). Then the transformation
τ : (n, m)7−→(2n−2m, n−m+ 1), n ≥m induces a map on the set of the solutions in each case.
Mapτ has a great importance with useful properties. For (n, m)∈A± letδ(n, m)∈Z denote the distance n−m of the exponents n and m.
Property 1. δ(τ(n, m)) = δ(n, m)−1. In particular,τ(n, m)6= (n, m), i.e. the map has no fixed points.
Property 2. If (n, m)∈R±\ {(0,1)}, more precisely if (n, m) = (2t, t+ 1), t ≥1, then τ(n, m) = (2t−2, t)∈R± is thelower neighbor solution of (n, m) in R±. Thus the elements of the set R± are ordered by τ. Moreover, δ(τ(2t, t+ 1)) = t−2 shows
that all integers at least -1 occur as a difference of the exponents in the solution of (1) if t≥1.
Property 3. If (n, m) is an exceptional solution, i.e. (n, m)∈ A±\R±, then τ(n, m) ∈ R±, respectively, sinceτ(n, m) = (2δ(n, m), δ(n, m) + 1).
By Properties 1-3 we have to determine the pairs (n, m)6= (n1, m1) such that τ(n, m) = τ(n1, m1), where either (n, m) or (n1, m1) is inR±, and n1 =n+d,m1 =m+dhold with some suitable positive integerd. In other words, we need to solve the system of equations
2n±α·2m+α2 = x2,
2n+d±α·2m+d+α2 = y2, (2)
respectively, in integers n ≥m ≥1,d≥1, and odd y > x≥1.
These equalities immediately imply
y2−α2 = 2d x2−α2
. (3)
Taking (2) modulo α we see that d must be even since n is even. Put d= 2t. In the next section, an algorithm will be developed for solving (3) with d= 2t.
We note that in this section we have concentrated on the main case of the problem, some other choices like m= 0 orn < m (in 2n−α·2m+α2 =x2) are not analyzed here.
Of course, the algorithm will be able to handle such eliminated specifications when we examine the equations 2n±α·2m+α2 =x2 separately.
3 Algorithm for solving y
2− α
2= 2
2t(x
2− α
2)
In this section, we assume only that 5 ≤ α ∈ P is fixed, and we omit the condition (2|p) =−1 on the Legendre symbol. We look for the solutions [t, x, y] to
y2−α2 = 22t(x2−α2) (4)
in the integers t≥1, and y > x > α. Obviously, (x, y) = 2t−1α, (22t−1−1)α
, t≥2 (5)
describes an infinite family of solutions. Hence our actual purpose is to determine the so-called sporadic solutions do not belong to this family.
3.1 Theoretical background
First we take two simple observations related to (4).
Observation 1. If [t, x, y] is a solution to (4), then y <2tx.
Indeed, y2 = 22tx2+α2(1−22t)<22tx2 impliesy <2t immediately.
Observation 2. Ift is fixed, then the solutions can be easily determined from the equiv- alent form
(22t−1)α2 = (2tx−y)(2tx+y) (6) of (4). Really, one must check only the positive divisors of tα = (22t−1)α2.
Now we distinguish two cases.
First suppose 2t> α. Thus α/2t <1, and by α≥ 5 we have t ≥3. Clearly, y is odd, therefore gcd(y−α, y+α) = gcd(y−α,2α) = 2 or 2α. Consequently, from
y−α
2 · y+α
2 = 22t−2(x2−α) we conclude
22t−2 | y−α
2 or 22t−2 | y+α 2 . Then
y = 22t−1k±α (7)
follows, where k is a positive integer. Combining (4) and (7) we deduce
22t−2k2±αk =x2−α2. (8)
In the − case y= 22t−1k−α <2tx (see Observation 1) leads immediately to 2t−1k− α/2t < x. Since now α/2t < 1 holds, we have 2t−1k ≤ x. If 2t−1k = x, then by (8) we gain k = α, and then x = 2t−1α. But this belongs to the infinite family. Consequently, x≥2t−1k+ 1.
In the + case, by Observation 1, 2tx > y = 22t−1k +α > 22t−1k follows, and then x≥2t−1k+ 1. This coincides the final conclusion of the minus case.
Combining (8) andx≥2t−1k+ 1, we obtainα2−1≥(2t∓α)k. The condition α <2t guarantees the positivity of 2t∓α. Thus 2t≤α2±α−1, and finally we have
t≤
log2(α2±α−1)
≤
log2(α2+α−1)
. (9)
From algorithmic point of view (9) is very advantageous, since it provides finitely many values for t, and according to Observation 2 these values can be handled.
Secondly, assuming 2t ≤ α, where t ≥ 1, it implies right away t ≤ blog2(α)c. Note that α≤α2±α−1 always holds for α≥5.
Hence we obtain the following
Corollary 1. Assume that α is fixed. Equation (4) possesses finitely many solutions [t, x, y], further t≤ blog2(α2+α+ 1)c.
3.2 The algorithm, and computational experiences
Input: α (5≤α ∈P).
Output: Set S of sporadic solutions [t, x, y] to (4) (t ≥1,y > x > α).
Procedure Pre:
[1] S :={}; tub :=blog2(α2+α−1)c.
[2] for t from 1 to tub do
[3] check (6) by Observation 2;
[4] add the solutions, if exist and sporadic, to S;
[5] end of for.
Table 3 collects the solution to (4) for 5≤α <37, α ∈P, (2|α) = −1.
α sporadic solutions [t, x, y] to (4)
5 [1,19,37], [1,7,11], [2,47,187], [2,16,61], [3,33,261], [3,6,27]
11 [1,91,181], [1,31,59], [2,227,907], [2,76,301], [2,46,179], [2,17,53], [3,159,1269]
[3,24,171], [4,193,3083], [4,14,139], [6,34,2059]
13 [1,127,253], [1,43,83], [2,317,1267], [2,106,421], [2,64,251], [2,23,77]
[3,222,1773], [3,33,243], [4,449,7181], [4,28,397]
19 [1,271,541], [1,91,179], [2,677,2707], [2,226,901], [2,136,539], [2,47,173]
[3,474,3789], [3,69,531], [4,959,15341], [4,58,877], [6,257,16403]
29 [1,211,419], [1,631,1261], [2,107,413], [2,316,1259], [2,526,2101], [2,1577,6307]
[3,159,1251], [3,1104,8829], [4,133,2077], [4,2234,35741], [5,146,4579]
[5,4481,143389], [6,769,49181]
Table 3: solution to (4) for 5≤α <37, α∈P, (2|α) =−1
In case of the largest prime α= 2796203<3·106 we found solution to (1) there exist 46 triples satisfy (4), the largest ones in t are
[t, x, y] = [22,3495254,8796095818411],[22,11728129622017,49191340986142993067].
(Now tub= 42.)
Remark 2. At the application of the solutions to (4) only the odd x values are used.
Hence the terms belonging to the infinite family x= 2t−1α (t ≥2) are not taken account of, furthermore we will make a selection among the sporadic solutions.
Looking Table 3 above,t = 1 appears twice in each row,t = 2 does four times except when α = 5, etc. These observations can be generalized for arbitrary α values with the given condition. In other words, one can consider (4) with fixed t and prime variable α.
For instance, if t = 1, then 3α2 = (2x−y)(2x+y) follows. The six divisors of 3α2 are 1<3< α <3α < α2 <3α2, which form 3 possible divisor pairs:
(2x−y,2x+y) = (1,3α2), (3, α2), (α,3α).
The first two pairs give the solutions (x, y) =
3α2+ 1
4 ,3α2−1 2
,
α2+ 3
4 ,α2−3 2
,
respectively (the third one leads to x = α, a contradiction). Clearly, here x < y are positive integers greater than α. It explains the number of solutions with t = 1. Using this information, we know in advance that S is never empty.
By similar way, assuming t= 2, first we have 15α2 = (4x−y)(4x+y). Suppose that α ≥7. Then the 12 divisors of 15α2 form 6 pairs, among them 4 lead to solution, namely
(x, y) =
15α2+z
8 ,15α2−z 2
, z ∈ {1,3,5,15}.
(Ifα= 5, then 15α2 = 3·53 has only 8 divisors, and from the 4 pairs only 2 give solution.) The maintenance of larger values t is analogous.
4 Algorithm for solving 2
n− α · 2
m+ α
2= x
2The investigation of
2n−α·2m+α2 =x2 (10)
is split into two parts.
First assume 2n−α·2m <0, or equivalentlyn≤m+a, wherea=blog2(α)c<log2(α).
Then
0≤x2 ≤2m(2a−α) +α2, where 2a−α is negative. Subsequently,
2m ≤ α2 α−2a,
and m is bounded by mb =blog2(α2/(α−2a))c<2 log2(α). For each value 0≤m ≤mb we can apply Lemma 1 with D=−α·2m+α2 to determinen and x in (10).
Suppose now n > m+a and m > mb. Note that mb > log2(α) (this can be easily seen). Thusxis odd in (10). Then, following the idea which describesτ transformation in Section 2, we arrive at the sporadic solutions of equation (4). Indeed, x is odd, therefore the elements of the infinite family (5) are irrelevant.
If [t, x, y] is a sporadic solution to (4), and 2 - x also fulfils, then we must check whether it gives a solution to (10) or not. To do this, first let m=ν2(x2−α2). Then put n =ν2(x2−α2+α·2m). If the valuesn,m, andxsatisfy (10), moreovern 6= 2(m−1), then we obtain an element ofE−. Contrary, ifn= 2(m−1), moreover 2n+2t−α·2m+2t+α2 =y2 also holds, then (n+ 2t, m+ 2t, y)∈E−.
The arguments above provide the following corollary.
Corollary 3. The set E− of exceptional solutions to (10) is finite for any fixed α.
4.1 The algorithm, and running consequences
Input: α (5≤α ∈P).
Output: Set E− of exceptional solutions (n, m, x)∈N3 to (10).
Procedure Eq.plus:
[1] E−:={}; a:=blog2(α)c; mb :=blog2(α2/(α−2a))c.
[2] for m from 0 to mb do [3] D:=α2−α·2m;
[4] solve 2n+D=x2 by Lemma 1;
[5] add the solutions, if exist and if not regular, to E−; [6] end of for.
[7] S:=Pre(α); nS := #S.
[8] for i from 1 to nS do
[9] let [t, x, y] be the ith element of S;
[10] if x is odd then m =:ν2(x2−α2); n =:ν2(x2−α2+α·2m);
[11] if 2n−α·2m+α2 =x2 then
[12] if (n, m, x) is exceptional then add it to E− endif;
[13] if (n, m, x) is regular then
[14] if 2n+2t−α·2m+2t+α2 =y2 then add (n+ 2t, m+ 2t, y) to E− endif;
[15] endif;
[16] endif;
[17] endif;
[18] end of for.
The algorithm is implemented in Maple, and we checked the possible α values up to 3·106. The computational results prove
Theorem 4. If the non-negative integers n, m, x and α with α ∈ P, (2|α) = −1, α <3·106 satisfy
2n−α·2m+α2 =x2,
then they are either listed up in Table 1 or (n, m, x) = (2t, t+ 1,|2t−α|).
Remark 5. The solution (n, m, x) = (6,3,7) for α= 3 was determined in [6].
5 Algorithm for solving 2
n+ α · 2
m+ α
2= x
2Recall that we intend to handle
2n+α·2m+α2 =x2 (11)
only with the condition n ≥ m. The principal reason is that transformation τ does not exist if n < m.
First assume n > m ≥ 1. Recall again the idea appears in Section 2 concerning transformation τ. It leads again to (4). Since x is odd in (11), therefore the elements of the infinite family (5) are irrelevant. If [t, x, y] is a sporadic solution to (4), and 2-xalso fulfils, then we need to check whether it gives a solution to (11) or not. Repeat the steps described in the minus case replacing − by + where necessary. Thus we obtain a subset of E+. In the sequel, we analyze the remaining specific cases.
Suppose now m = 0. For a given α let D = α +α2, and use Lemma 1 to solve 2n+D=x2. Merge the solutions to E+ if exist.
Finally, considern =m ≥1 which provides
(α+ 1)·2n+α2 =x2. (12)
Ifn = 1 we obtain (α+ 1)2+ 1 =x2, a contradiction byα ≥5. The casen= 2 belongs to the infinite family n = 2t, m=t+ 1 with t = 1. Thus n≥3 can be assumed. Therefore x must be odd, furthermore (2|p) =−1 implies α= 8β±3. Hence α+ 1 = 4(2β+ 1) or 2(4β−1).
Rewrite (12) into the form
2n−2+ε α+ 1
2ε = x−α
2 · x+α 2 ,
where ε = ν2(α+ 1) = 1 or 2. Since α - x, we conclude gcd((x−α)/2,(x+α)/2) = 1.
Subsequently, 2n−2+ε divides exactly one of (x−α)/2 and (x+α)/2. Therefore
x= 2n−1+εk±α (13)
holds for some suitable positive odd integer k. Combining it and (12), we have α+ 1
2ε = 2n−2+εk2±αk. (14)
The + case is not possible since (α+ 1)/2ε = 2n−2+εk2+αk > α contradicts to α≥5.
In the−case 2n−2+εk2−αk ≤(2n−2+ε−α)k2, and we distinguish two options according to the sign of 2n−2+ε−α. Assuming 2n−2+ε−α <0, it leads ton≤nb := 2−ε+blog2(α)c.
This gives finitely many possibilities for n in (12). Thus one must check if xis integer. If 2n−2+ε−α >0, then
k ≤ s
α+ 1
2ε(2n−2+ε−α) ≤kb :=
rα+ 1 2ε .
Verifying the appropriate range of k, we may get values forn by (14). Then (13) provides some integers x. If such an xvalue satisfies (12), then we obtain a new element of E+.
Thus we conclude
Corollary 6. The set E+ of exceptional solutions with n ≥ m to (11) is finite for any fixed α.
5.1 The algorithm, and running experiments
Input: α (5≤α ∈P).
Output: Set E+ of exceptional solutions (n, m, x)∈N3 to (11).
Procedure Eq.plus:
[1] E+:={}; D:=α2+α;
[3] solve 2n+D=x2 by Lemma 1;
[4] add the solutions, if exist and if not regular, to E+; [5] ε:=ν2(α+ 1); nb := 2−ε+blog2(α)c;
[6] for n from 3 to nb do [7] if x:=p
(α+ 1)·2n+α2 is integer then add (n, n, x) to E+ endif;
[8] end of for.
[9] kb :=p
(α+ 1)/2ε;
[10] for k from 1 by 2 to kb do
[11] n:= 2−ε+blog2((α+ 1)/(2εk2) +α/k)c; x:= 2n−1+εk−α;
[12] if 2n(α+ 1) +α2 =x2 then add (n, n, x) to E+ endif;
[13] end of for.
[14] S:=Pre(α); nS := #S.
[15] for i from 1 to nS do
[16] let [t, x, y] be the ith element of S;
[17] if x is odd then m =:ν2(x2−α2); n =:ν2(x2−α2−α·2m);
[18] if 2n+α·2m+α2 =x2 then
[19] if (n, m, x) is exceptional then add it to E+ endif;
[20] if (n, m, x) is regular then
[21] if 2n+2t+α·2m+2t+α2 =y2 then add (n+ 2t, m+ 2t, y) to E+ endif;
[22] endif;
[23] endif;
[24] endif;
[25] end of for.
The algorithm is implemented in Maple, and we checked the possible α values up to 3·106. The computational results prove
Theorem 7. If the non-negative integers n, m, x andα with n≥m, α∈P, (2|α) = −1, α <3·106 satisfy
2n+α·2m+α2 =x2,
then they are either listed up in Table 2 or (n, m, x) = (2t, t+ 1,2t+α).
Remark 8. For α= 3 the three solutions above were determined in [6].
Acknowledgment
The author is grateful to the anonymous referees for the valuable remarks. The re- search was supported by Hungarian National Foundation for Scientific Research Grant No. 128088, and by Faculty of Economics, University J. Selye.
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