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Article

Linear Relations between Numbers of Terms and First Terms of Sums of Consecutive Squared Integers Equal to Squared Integers

by
Vladimir Pletser
1,2
1
European Space Research and Technology Centre, European Space Agency, 2201 AZ Noordwijk, The Netherlands
2
Blue Abyss, Pool Innovation Centre, Cornwall TR15 3PL, UK
Symmetry 2024, 16(2), 146; https://doi.org/10.3390/sym16020146
Submission received: 29 December 2023 / Revised: 14 January 2024 / Accepted: 24 January 2024 / Published: 26 January 2024
(This article belongs to the Section Mathematics)

Abstract

:
The classical problem of finding all integers a and M such that the sums of M consecutive squared integers a + i 2 equal the squared integer s 2 , where M is the number of terms in the sum, a 2 the first term and a 1 , 0 i M 1 , yields remarkable regular linear features when plotting the values of M as a function of a. These linear features correspond to groupings of pairs of a values for successive same values of M found on either side of straight lines of equation μ M = 2 a + c , where c is an integer constant and μ a parameter taking some rational values, called allowed values. We find expressions of a and s as a function of M for the allowed values of μ and M and parametric expressions of a, M, and s. Further, Pell equations deduced from the conditions of M are solved to find the allowed values of μ and to provide all solutions in a and M. These results yield new insights into the overall properties of the classical problem of the sums of consecutive squared integers equal to squared integers and allow us to solve this problem completely by providing all solutions in infinite families.

1. Introduction

Throughout this paper, we are considering sums of squared integers equal to another squared integer
i = 0 M 1 a + i 2 = s 2
where M , a , i , s are positive integers, M is the number of terms in the sum and a 2 is the first term in the sum.
The study of integer squares equal to sums of consecutive squared integers can be dated back to 1873, when Lucas stated [1] that 1 2 + + n 2 is an integer square only for n = 1 and 24. Lucas proposed later in 1875 [2] the well-known cannonball problem. This problem can clearly be written as a Diophantine equation i = 1 M i 2 = M M + 1 2 M + 1 / 6 = s 2 . The only solutions are s 2 = 1 and 4900, which correspond to the sum of the first M squared integers for M = 1 and M = 24 . This was partially proven by Moret-Blanc [3] and Lucas [4], and entirely proven later on by Watson [5] (using elliptic functions for one case), Ljunggren [6], Ma [7] and Anglin [8].
Instead of starting at 1, finding all integers a 1 for which the sum of M consecutive integer squares starting from a 2 is itself an integer square s 2 is a more general problem. Various approaches have been followed in attempts at finding solutions to this problem. Alfred proposed [9] several necessary conditions on M (with the notations of this paper), yielding M = 2 , 11 , 23 , 24 , 26 , until M = 500 by solving congruence equations of M, without conclusions for M = 107 , 193 , 227 , 275 , 457 . Philipp [10] showed further that solutions exist for M = 107 , 193 , 457 but not for M = 227 , 275 and proved that there are a finite or an infinite number of solutions depending on whether M is or is not a square integer. Laub demonstrated [11] that the set of M yielding the sum of M consecutive squared integers being a squared integer is infinite and has density zero. Eight necessary conditions were imposed on M by Beeckmans [12], yielding a list of values of M < 1000 with the corresponding smallest value of a > 0 . Two cases for M = 25 and 842 were found satisfying the eight necessary conditions but not yielding solutions to the problem.
This author showed [13] that (1) has no integer solutions if M 3 , 5 , 6 , 7 , 8 or 10 m o d   12 and that (1) has integer solutions for non-squared integer M congruent to 0 , 9 , 24 or 33 m o d   72 , or to 1 , 2 or 16 m o d   24 , or to 11 m o d   12 , and for squared integer M congruent to 1 m o d   24 .
In this paper, we investigate and characterize the properties of groupings of pairs of integers a for same values of M. These groupings are found in an a , M plot on either side of the inclined straight lines of equation μ M = 2 a + c , where c is a constant, that can be approximated by μ M 2 a , for rational values of μ = η / δ , where η , δ are integers such that μ is an irreducible fraction.
In Section 2.1, we give a definition of pairs of integers a solutions of (1) for the same value of M and a theorem giving the expressions of a and s as a function of M for allowed values of μ and M. In Section 2.2, we give parametric expressions of a, M and s. In Section 2.3, we solve Pell equations to find the allowed values of μ . In Section 3, we give and discuss examples of some infinite families of solutions of (1). Conclusions are drawn in Section 4.

2. Materials and Methods

2.1. Observed Linear Features in the (a,M) Plot

For M > 1 , a , i , s Z * , the sum of M consecutive squared integers a + i 2 equaling a squared integer s 2 can be written [14] as
i = 0 M 1 a + i 2 = M a + M 1 2 2 + M 2 1 12
For 1 a 10 5 and 2 M 10 5 , there are only 4078 couples of values of a and M among the approximately 10 10 possibilities such that (2) holds. Figure 1 shows the distribution of these 4078 couples in an a , M plot, where several groupings of interest are seen.
The most visible is the grouping around a straight line of equation M = 2 a + c , where c is a constant, corresponding to a double infinite family of integers a. This grouping starts with the identity and the Pythagorean relations
0 2 + 1 2 = 1 2
3 2 + 4 2 = 5 2
for the same value of M = 2 and, respectively, for a = 0 and 3. This double infinite family has the property that couples of a values correspond to the same value of M. There are other similar groupings and double infinite families around straight lines of general equation μ M = 2 a + c μ for certain rational values of μ > 0 , and where c μ is a constant different for each value of μ . Only groupings around inclined lines are considered, as the limit cases of μ = 0 and μ corresponding to groupings around, respectively, vertical and horizontal lines are not treated here. The “horizontal” case for which one or several solutions in a exist for each value of M was investigated in [13,14].

2.2. Definition of Pairs and Theorem on Conditions on μ, M and a

Definition 1.
For 1 j 2 , M μ , k > 1 , a j , μ , k Z * , for a given μ Q + , two integers a j , μ , k are called a pair a 1 , μ , k , a 2 , μ , k if for the same value of M μ , k and k Z , the relations
a 1 , μ , k + a 2 , μ , k = μ M μ , k + 1
a 2 , μ , k a 1 , μ , k = f μ , k
hold, where f μ , k = f μ k is a linear integer function of k for each value of μ, yielding
μ M μ , k = 2 a j , μ , k ± f μ , k 1
where the upper or lower sign is taken for j = 1 or 2.
The two families of a 1 , μ , k and a 2 , μ , k are characterized for each value of μ around the straight line of equation μ M = 2 a + c μ in the following theorem. However, relations (5) to (7) hold only for certain values of M μ , k and of μ , called allowed values, which are determined further.
Theorem 1.
For 1 j 2 , i , η , δ , M μ , k > 1 , a j , μ , k Z * , k , s j , μ , k Z , for allowed values of μ Q + , let μ = η / δ be an irreducible fraction; if a 1 , μ , k , a 2 , μ , k is a pair of integers  a j , μ , k  for the same value of  M μ , k  and if
M μ , k = 3 f μ , k 2 1 3 μ + 1 2 + 1 = δ 2 3 f μ , k 2 1 3 η + δ 2 + δ 2
holds k Z , then the sums of squares of M μ , k consecutive integers a j , μ , k + i for i = 0 to M μ , k 1 are always equal to squared integers s j , μ , k 2 , with
a j , μ , k = 1 2 δ η M μ , k + δ M μ , k 3 η + δ 2 + δ 2 + δ 2 3
s j , μ , k = M μ , k 2 δ M μ , k 3 η + δ 2 + δ 2 + δ 2 3 η + δ
where the upper (respectively, lower) sign is taken for j = 1 (respectively, 2).
Proof. 
Let 1 j 2 , i , η , δ , M μ , k > 1 , a j , μ , k Z * , k , s j , μ , k Z , μ = η / δ Q + forming an irreducible fraction, i.e., gcd η , δ = 1 . Let further f μ , k be a yet-unknown integer function of k for each value of μ . Replacing in the second equality of (2) M by M μ , k and a by a j , μ , k from (7) yields, successively,
i = 0 M μ , k 1 a j , μ , k + i 2 = M μ , k 4 2 a j , μ , k + M μ , k 1 2 + M μ , k 2 1 3 = M μ , k 4 μ + 1 M μ , k f μ , k 2 + M μ , k 2 1 3 = M μ , k 2 4 3 η + δ 2 + δ 2 M μ , k 3 δ 2 + 3 f μ , k 2 1 3 M μ , k 2 η + δ δ f μ , k ]
where the upper (respectively, lower) sign in (11) is taken for j = 1 (respectively, 2). For the expression between brackets in (11) to be a square, replace in (11) f μ , k by
f μ , k = M μ , k 3 η + δ 2 + δ 2 + δ 2 3 δ 2
from (8), immediately yielding (10). Replacing f μ , k (12) in (7) then yields (9). □
In addition, from Theorem 2, the following relations hold k Z
a 2 , μ , k a 1 , μ , k = M μ , k 3 η + δ 2 + δ 2 + δ 2 3 δ 2
s 2 , μ , k + s 1 , μ , k = M μ , k M μ , k 3 η + δ 2 + δ 2 + δ 2 3 δ 2
= M μ , k a 2 , μ , k a 1 , μ , k
s 2 , μ , k s 1 , μ , k = M μ , k η + δ δ
= a 2 , μ , k + a 1 , μ , k + M μ , k 1

2.3. Parametric Expressions of f μ , k , M μ , k , a j , μ , k , s j , μ , k

The above results hold only for certain values of M μ , k and of μ Q + , which are called allowed values. They can be determined as follows. Relation (8) also reads
δ f μ , k 2 M μ , k η + δ 2 = δ 2 M μ , k + 1 3
It was shown [15] that for (18) to hold:
-
δ 0 m o d   6 and η 1 or 5 m o d   6 , M μ , k 0 or 24 m o d   72 and M μ , k m o d   δ 2 / 3 0 ;
-
δ 1 or 5 m o d   6 and η 1 , 3 or 5 m o d   6 and either f μ , k 1 m o d   2 and M μ , k 2 m o d   24 or f μ , k 0 m o d   2 and M μ , k 11 m o d   12 , and M μ , k m o d   δ 2 0 .
Parametric expressions of f μ , k , M μ , k , a j , μ , k and s j , μ , k as a function of k Z , μ = η / δ and initial values are found as follows.
Theorem 2.
For 1 j 2 , η , δ , M μ , k > 1 Z + , a j , μ , k Z * , k , s j , μ , k Z , μ , ν Q + , for allowed values of μ = η / δ and for pairs a 1 , μ , k , a 2 , μ , k , f μ , k is a linear function of k, M μ , k and  a j , μ , k  are quadratic functions of k and  s j , μ , k  is a cubic function of k, as follows:
f μ , k = 3 η + δ 2 + δ 2 ν k + f μ , 0
M μ , k = 3 δ 2 3 η + δ 2 + δ 2 ν 2 k 2 + 6 δ 2 f μ , 0 ν k + M μ , 0
a j , μ , k = 1 2 3 η δ 3 η + δ 2 + δ 2 ν 2 k 2 + 6 η δ f μ , 0 3 η + δ 2 + δ 2 ν k +   a j , μ , 0
s j , μ , k = 1 2 3 δ 2 3 η + δ 2 + δ 2 2 ν 3 k 3 +   3 δ 3 η + δ 2 + δ 2 3 δ f μ , 0 η + δ ν 2 k 2 + 3 δ f μ , 0 η + δ 2 η + δ 2 + δ 2 ν k + s j , μ , 0
where ν = 1 for δ 1 or 5 m o d   6 , ν = 2 / 3 for δ 0 m o d   6 and where the upper (respectively, lower) sign is taken for j = 1 (respectively, 2).
Proof. 
For 1 j 2 , η , δ , x , M μ , k > 1 Z + , a j , μ , k Z * , k , s j , μ , k Z , μ , ν Q + , for allowed values of μ = η / δ and for pairs a 1 , μ , k , a 2 , μ , k , let f μ , k (6) be a linear function of k, f μ , k = x k + f μ , 0 , where f μ , 0 = a 2 , μ , 0 a 1 , μ , 0 is the initial value for k = 0 of the difference (6) and x an integer function to be defined for some parameters. Then, (8) yields
M μ , k = δ 2 3 x k + f μ , 0 2 1 3 η + δ 2 + δ 2 = 3 δ 2 x k x k + 2 f μ , 0 3 η + δ 2 + δ 2 + δ 2 3 f μ , 0 2 1 3 η + δ 2 + δ 2
The second term on the right of (23) is M μ , 0 by (8).
(i) If δ 1 or 5 m o d   6 , as M μ , k Z + , 3 η + δ 2 + δ 2 must divide x in the first term of (23), then yielding (19) and (20) with ν = 1 .
(ii) If δ 0 m o d   6 , simplifying the first term by 3, (23) reads
M μ , k = δ 2 x k x k + 2 f μ , 0 η + δ 2 + δ 2 / 3 + M μ , 0
As M μ , k Z + , η + δ 2 + δ 2 / 3 is a factor of x. However, as η 1 or 5 m o d   6 for δ 0 m o d   6 , η + δ 2 + δ 2 / 3 1 m o d   2 and as f μ , k m o d   2 f μ , 0 m o d   2   k Z , x must be replaced by 2 η + δ 2 + δ 2 / 3 , then yielding (19) and (20) with ν = 2 / 3 .
(iii) Further, replacing f μ , k and M μ , k by (19) and (20) in a j , μ , k from (7) and in s j , μ , k (10) with (12) yields directly (21) and (22) with the upper (or lower) sign for j = 1 (or 2). □

2.4. Finding Allowed Values of μ = η / δ and M μ , k

Finding the allowed values of μ = η / δ , M μ , 0 and f μ , 0 requires solving the generalized Pell Equation (18) for k = 0 in variables δ f μ , 0 and η + δ .
In general, for X , Y , D , N , x f , y f , n Z + and D square-free (i.e., D Z ), a generalized Pell equation X 2 D Y 2 = N admits either no solution or one or several fundamental solution(s) X 1 , Y 1 and also one or several infinite branches of solutions X n , Y n . Several methods exist to find the fundamental solutions of the generalized Pell equation (see [16,17,18]). Two methods are used further: first, a brute-force search method, i.e., trying several values of Y until the smallest X 1 = N + D Y 1 2 Z + is found; second, Matthews’ method [19] based on an algorithm by Frattini [20,21,22] using Nagell’s bounds [23,24]. Once fundamental solution(s) X 1 , Y 1 have been found one way or another, noting x f , y f , the fundamental solutions of the related simple Pell equation X 2 D Y 2 = 1 , the other solutions X n , Y n can be found by
X n + D Y n = ± X 1 + D Y 1 x f + D y f n
for a proper choice of sign ± [25], which can also be written as a function of Chebyshev’s polynomials [14]
X n = X 1 T n 1 x f + D Y 1 y f U n 2 x f
Y n = X 1 y f U n 2 x f + Y 1 T n 1 x f
where T n 1 x f and U n 2 x f are Chebyshev polynomials of the first and second kinds evaluated at x f .
The generalized Pell Equation (18) can be written as
λ f μ , 0 2 λ 2 M μ , 0 δ 2 η + δ 2 = λ 2 M μ , 0 + 1 3
with X = λ f μ , 0 , Y = η + δ , D = λ 2 M μ , 0 / δ 2 and N = λ 2 M μ , 0 + 1 / 3 , and where λ = 1 if δ 1 or 5 m o d   6 and λ = 3 if δ 0 m o d   6 .
To use Matthews’ method [19], the parameters D and N must be fixed with values of δ and M μ , 0 that can be chosen from the allowed congruent values (see Section 2.3) and be tried one by one until fundamental solutions are found. Alternatively, fixing the values of η and δ , a brute-force search method can be used to find f μ , 0 Z + for the smallest value of M μ , 0 Z + , with, from (12),
f μ , 0 = M μ , 0 3 η + δ 2 + δ 2 + δ 2 3 δ 2
Relation (29) then yields the allowed values of μ , M μ , 0 and f μ , 0 given in Table 1 for δ = 1 , μ = η Z + , for 0 μ 100 ([26] gives all values of μ + 1 such that 3 μ + 1 2 + 1 is prime, for which (29) holds.) and in [27] for δ 1 , μ = η / δ Q + , for 0 < η , δ 100 .
Once a set of values has been found for η , δ , M μ , 0 and f μ , 0 as fundamental solution(s) of the generalized Pell Equation (28), other allowed values for η and f μ , 0 can be found from the other solutions of (28) using the values of δ and M μ , 0 with (26) and (27), written as
f μ n , 0 = f μ 1 , 0 T n 1 x f + λ M μ , 0 δ 2 η 1 + δ y f U n 2 x f
η n = λ f μ 1 , 0 y f U n 2 x f + η 1 + δ T n 1 x f δ

3. Results and Discussion

To demonstrate the soundness of the method exposed above using the Pell equation approach, we give here several examples.
Example 1.
For δ = 1 and μ = η = 1 , M 1 , 0 = 2 and f 1 , 0 = 3 from Table 1. Using M 1 , 0 and δ as constants in (28) with λ = 1 , it reduces to a simple Pell equation f μ , 0 2 2 μ + 1 2 = 1 (see, e.g., [28,29,30,31]) that admits the single fundamental solution X 1 , Y 1 = f μ 1 , 0 , μ 1 + 1 = 3 , 2 or f μ 1 , 0 , μ 1 = 3 , 1 and an infinity of other solutions that can be found n Z + by
f μ n , 0 = 3 + 2 2 n + 3 2 2 n 2 = 3 , 17 , 99 , 577 , 3363 ,
μ n = 3 + 2 2 n 3 2 2 n 2 2 1 = 1 , 11 , 69 , 407 , 2377 ,
where μ n are the Pell numbers [32] of even indices minus one. These new values of f μ n , 0 , μ n for n > 1 define new groupings around straight lines of general equation μ n M = 2 a + c μ n , with the initial value M μ n , 0 = 2 .
  • For δ = 1 , η = 1 , i.e., μ = 1 , M 1 , 0 = 2 , f 1 , 0 = 3 , ν = 1 , (20) to (22) yield
  • M 1 , k = 39 k 2 + 18 k + 2 ,
  • a 1 , 1 , k = 39 k 2 + 5 k / 2 , a 2 , 1 , k = 39 k 2 + 31 k + 6 / 2 ,
  • s 1 , 1 , k = 507 k 3 + 273 k 2 + 44 k + 2 / 2 , s 2 , 1 , k = 507 k 3 + 429 k 2 + 116 k + 10 / 2 ,
  • and values of M μ , k , a 1 , μ , k and a 2 , μ , k for 10 k 10 are given in Table 2.
Figure 2 shows the distribution of M versus a for μ = 1 .
Example 2.
For δ = 1 and μ = η = 5 , M 5 , 0 = 11 and f 5 , 0 = 20 from Table 1. Using M 5 , 0 and δ as constants in (28) with λ = 1 yields the generalized Pell equation f μ , 0 2 11 μ + 1 2 = 4 . Using Matthews’ method [19] yields the single fundamental solution f μ 1 , 0 , μ 1 + 1 = 2 , 0 , which is of no use. However, as the right-hand term is a squared integer, the equation can be rewritten as a simple Pell equation f μ , 0 / 2 2 11 μ + 1 / 2 2 = 1 , which admits the fundamental solution f μ 1 , 0 / 2 , μ 1 + 1 / 2 = 10 , 3 or f μ 1 , 0 , μ 1 = 20 , 5 and an infinity of other solutions n Z +
f μ n , 0 = 10 + 3 11 n + 10 3 11 n = 20 , 398 , 7940 , 158402 ,
μ n = 10 + 3 11 n 10 3 11 n 11 1 = 5 , 119 , 2393 , 47759 ,
  • For δ = 1 , η = 5 , i.e., μ = 5 , M 5 , 0 = 11 , f 5 , 0 = 20 , ν = 1 , (20) to (22) yield (see Table 2)
  • M 5 , k = 327 k 2 + 120 k + 11 ,
  • a 1 , 5 , k = 1635 k 2 + 491 k + 36 / 2 , a 2 , 5 , k = 1635 k 2 + 709 k + 76 / 2 ,
  • s 1 , 5 , k = 35643 k 3 + 17658 k 2 + 2879 k + 154 / 2 ,
  • s 2 , 5 , k = 35643 k 3 + 21582 k 2 + 4319 k + 286 / 2 .
Figure 3 shows the distribution of M versus a for μ = 5 .
Example 3.
For η = 1 and δ = 6 , i.e., μ = 1 / 6 , M 1 / 6 , 0 = 312 and f 1 / 6 , 0 = 23 [27]. Using M 1 / 6 , 0 and δ as constants in (28) with λ = 3 yields 3 f μ , 0 2 78 η + 6 2 = 939 , which, by [19], has two fundamental solutions 3 f μ 1 , 0 , η 1 + 6 = 69 , 7 and 381 , 43 , yielding f μ 1 , 0 , η 1 = 23 , 1 and 127 , 37 , respectively. The fundamental solutions of the related simple Pell equation X 2 78 Y 2 = 1 are x f , y f = 53 , 6 . Other values of f μ n , 0 , η n can be found on the two infinite branches corresponding to these two fundamental solutions by (30) and (31) as
f μ n , 0 = 23 T n 1 53 + 1092 U n 2 53
= 23 , 2311 , 244943 , 25961647 , 2751689639 ,
η n = 414 U n 2 53 + 7 T n 1 53 6
= 1 , 779 , 83197 , 8818727 , 934702489 ,
for the first fundamental solution and
f μ n , 0 = 127 T n 1 53 + 6708 U n 2 53
= 127 , 13439 , 1424407 , 150973703 , 16001788111 ,
η n = 2286 U n 2 53 + 43 T n 1 53 6
= 37 , 4559 , 483841 , 51283211 , 5435537149 ,
for the second fundamental solution. For  η = 1 ,  δ = 6 , i.e.,  μ = 1 / 6 ,  M 1 / 6 , 0 = 312 ,  f 1 / 6 , 0 = 23 ,  ν = 2 / 3 , (20) to (22) yield (see Table 2).
  • M 1 / 6 , k = 8784 k 2 + 3312 k + 312 ,
  • a 1 , 1 / 6 , k = 732 k 2 + 215 k + 15 , a 2 , 1 / 6 , k = 732 k 2 + 337 k + 38 ,
  • s 1 , 1 / 6 , k = 535824 k 3 + 297924 k 2 + 55188 k + 3406 ,
  • s 2 , 1 / 6 , k = 535824 k 3 + 308172 k 2 + 59052 k + 3770 .
Figure 4 shows the distribution of M versus a for μ = 1 / 6 .

4. Conclusions

Definition 1 and Theorem 2 allowed us to show that regular linear features exist in the distribution of couples of values a and M in the a , M plot, where a 2 and M are the first term and the number of terms in sums of consecutive squared integers equal to integer squares. These regular features correspond to groupings of pairs of a values for successive same values of M found on either side of straight lines of equation μ M = 2 a + c for positive rational values of μ = η / δ and where c is a constant.
For allowed values of η and δ such as η 1 m o d   2 and δ 0 , 1 or 5 m o d   6 , if M μ , k = δ 2 3 a 2 , μ , k a 1 , μ , k 2 1 / 3 η + δ 2 + δ 2 holds k Z and for pairs a 1 , μ , k , a 2 , μ , k , then the sums of M μ , k consecutive squared integers starting with a 1 , μ , k or a 2 , μ , k are always equal to squared integers s 1 , μ , k 2 or s 2 , μ , k 2 k Z . Parametric equations are found as a function of k Z : linear for f μ , k = a 2 , μ , k a 1 , μ , k , quadratic for M μ , k , a 1 , μ , k and a 2 , μ , k and cubic for s 1 , μ , k and s 2 , μ , k .
The allowed values of η , δ , M μ , 0 and of the difference f μ , 0 = a 2 , μ , 0 a 1 , μ , 0 are found by solving the generalized Pell equation δ f μ , 0 2 M μ , 0 η + δ 2 = δ 2 M μ , 0 + 1 / 3 and further allowed values of η n and f μ n , 0 can be calculated for fixed values of δ and M μ , 0 using Chebyshev polynomials.
This approach allows us to completely solve the classical problem of the sums of consecutive squared integers equal to squared integers and provide all solutions in infinite families.

Funding

This research received no external funding.

Data Availability Statement

There are no data associated with this work.

Acknowledgments

This research was conducted under the good auspice of the European Space Agency Technical and Research Centre (The Netherlands) and the Engineering and Technology Centre for Space Utilization (Beijing).

Conflicts of Interest

The author declares no conflicts of interest.

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Figure 1. Distribution of M versus a for the 4078 couples a ,   M found such as (2) holds, with 1 a 10 5 and 2 M 10 5 .
Figure 1. Distribution of M versus a for the 4078 couples a ,   M found such as (2) holds, with 1 a 10 5 and 2 M 10 5 .
Symmetry 16 00146 g001
Figure 2. Distribution of M versus a for μ = 1 .
Figure 2. Distribution of M versus a for μ = 1 .
Symmetry 16 00146 g002
Figure 3. Distribution of M versus a for μ = 5 .
Figure 3. Distribution of M versus a for μ = 5 .
Symmetry 16 00146 g003
Figure 4. Distribution of M versus a for μ = 1 / 6 .
Figure 4. Distribution of M versus a for μ = 1 / 6 .
Symmetry 16 00146 g004
Table 1. Values of M μ , 0 and f μ , 0 for 0 μ 100 .
Table 1. Values of M μ , 0 and f μ , 0 for 0 μ 100 .
μ M μ , 0 f μ , 0 μ M μ , 0 f μ , 0 μ M μ , 0 f μ , 0
12329261536332633656
51120332995886795636650
774693547978869299
1121739139114927774671
1519422343593388112022843
19122221494911108831461015
2198369053173922528518743723
255065855538310969723470
27471925723272798
Table 2. Values of M μ , k , a 1 , μ , k , a 2 , μ , k for μ = 1 , 5 , 1 / 6 and 10 k 10 .
Table 2. Values of M μ , k , a 1 , μ , k , a 2 , μ , k for μ = 1 , 5 , 1 / 6 and 10 k 10 .
μ = 1 μ = 5 μ = 1 / 6
k M 1 , k a 1 , 1 , k a 2 , 1 , k M 5 , k a 1 , 5 , k a 2 , 5 , k M 1 / 6 , k a 1 , 1 / 6 , k a 2 , 1 / 6 , k
02031118383121538
−1231772185905015784532433
159223845810811210124089621107
−212273501079279725992882425132292
2194831121559377940174207233733640
−32991681322594663963326943259585615
34071832253314811284598930472487637
−4554302253476312116117001276081086710402
4698322377572314080145361541041258713098
−5887475413758619228187032033521724016653
51067500568878621683222482364721939020023
−612986876121106327975273412966642507724368
615147177981250330921315953364082765728412
−717879388501519438357376144075443437833547
7203997310671687441794425774539123738838265
−82354122811271997950374495225359924514344190
82642126813752189954302551945889844858349582
−92999155714432541864026630656820085737256297
93323160217222757868445694467416246124262363
−103722192517983151179313782438455927106569868
104082197521083391184223853339118327536576608
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Pletser, V. Linear Relations between Numbers of Terms and First Terms of Sums of Consecutive Squared Integers Equal to Squared Integers. Symmetry 2024, 16, 146. https://doi.org/10.3390/sym16020146

AMA Style

Pletser V. Linear Relations between Numbers of Terms and First Terms of Sums of Consecutive Squared Integers Equal to Squared Integers. Symmetry. 2024; 16(2):146. https://doi.org/10.3390/sym16020146

Chicago/Turabian Style

Pletser, Vladimir. 2024. "Linear Relations between Numbers of Terms and First Terms of Sums of Consecutive Squared Integers Equal to Squared Integers" Symmetry 16, no. 2: 146. https://doi.org/10.3390/sym16020146

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