Mathematical research grows from observations. One famous example is:
(1) 32 + 42 = 52
A magazine article led me to reference [1] where two similar relationships:
(2) 102 + 112 + 122 = 132 +
142 = 365
(3) 212 + 222 + 232 + 242
= 252 + 262 + 272
and some questions appear. The magazine item, the Brain Ticklers column
in [2], puts the basic issue this way:
We are interested in finding a sequence of 2n + 1 consecutive
positive integers, such that the sum of the squares of the first n + 1
integers equals the sum of the squares of the last n integers. The
simplest such sequence is 32 + 42 = 52.
(4) Give a formula in terms of n for the first integer in such a sequence. (The italics and numbering are added here.)
(*) Linton [3] pointed out that the first integer in such equations is
always n(2n + 1) when n is the number of terms on the right hand side of
the equation.
Gardner further reveals that the five squares that sum to
an adjacent four, and the six that sum to an adjacent five are,
Beldon [4] asserts that "any such sequence will lead to a quadratic
... that there is always a set of solutions symmetrical about zero ...
that the constant term in ... quadratic is always negative ... that
positive integral solutions always exist ... (and that
A general approach to this problem, viz the solution of
Sum[0 to n - 1] (a + n)2 = Sum[n to 2n - 2] (a + n)2
... leads to the quadratic equation:
a2 - 2 (n - 1)2a - (n - 1)2(2n - 1) = 0"
Item [4] concludes: Editorial Note A simpler soution is
found from the equation:
Sum[r = 0 to n] (a -r)2 = Sum[r = 1 to n] (a+r)2
which leads to the linear equation
a = 4 Sum n = 2n(n + 1)
and a knowledge of Sum n2 is not involved."
Write the editor's equation with the number of terms on the right hand
side set to 3 (value n in the above next to last equation). Algebraic simplification yields a = 24, where a is the number to be squared in left hand side's largest term.
Repeat for the case where there are 5 terms on the right, to see that 60 results.
If the first integer on the left hand side is n(2n +1), in the editor's
equation notation it is (a - n), where there are n terms on the right,
(n + 1) on the left. Then the last integer to square on the left is just a, and since
n + n(2n + 1) = 2n(n + 1), the editor's comment is consistent with
Linton.
Investigations and Exercises
1. Only (2) is evaluated. Give the sums for (1) and (3).
2. Find the sums for (5) and (6).
3. Describe the relationship between n, the number of terms on the right
hand side of the equation, and the overall sum, if there is a regular
rule. If there is none, show why that is so.
4. How can you relate these sums: 365, 2030, and 7230; to one another?
5. From (*), [n + 1]-many consecutive squares on the left and n-many on
the right, derive the general expressions:
(5.a) [n(2n + 1)]2 + [n(2n + 1) + 1]2 + ... +
[n(2n + 1) + n]2 = [n(2n + 1) + n + 1]2 +[n(2n + 1)
+ n + 2]2 + ... +
[n(2n + 3)]2
(5.b) [n(2n + 1)]2 + [n(2n + 1) + 1]2 + ... +
[2n(n + 1)]2 = [n(2n + 1) + n + 1]2 +[n(2n + 1)
+ n + 2]2 + ... +
[n(2n + 3)]2
6. Using, the results of 5., n = 1 as the base case, and mathematical induction,
show that that there are infinitely many 2n + 1 member sets of adjacent numbers
satisfying the property that sum of the squares of the lower n + 1
values equals the sum of the squares of the upper n.
Explanation
By writing out the result sought in general form and then doing
algebraic manipulation one can establish the following:
1. LHS has a^2 as its greatest term;
2. LHS remaining terms are (a - j)^2 for j= 1, 2, ... , n;
3. RHS has the n terms (a + j)^2 for j= 1, 2, ... , n;
4. The result* is that a = 4 Sum[ k, {k, 1, n}]
Hence for n =1 Sum is 1, a =4,
32 + 42 = 52;
... n =2 Sum is 3, a =12,
102 + 112 + 122 = 132 +
142;
... n =3 Sum is 6, a =24,
212 + 222 + 232 + 242
= 252 + 262 + 272 ;
... n =4 Sum is 10, a= 40, SEE (5) above
... n =5 Sum is 15, a= 60, SEE (6) above
etc., etc., etc.
Item 4. above is valid for any n, indicating that there are infinitely many
consecutives-squared whose sum follows the
pattern of the n =1 case (one more terms on the
LHS than on the RHS; all integers that are
squared being consecutive).
*Simply expand the squares of (a - j) and of (a
+ j) and simplify: only the cross-product
elements -2aj (LHS) and +2aj (RHS) remain. When
collected we get 4aj. Adding them up as j runs
from 1 through n yields (since we can divide
through eliminating a) 4 times the sum of the
integers from 1 to n.
References
1. Gardner, Martin, The Numerology of Dr. Matrix, NY: Simon and
Schuster, 1967.
2. Brain Ticklers, The Bent of Tau Beta Pi, Spring 2004.
3. Linton, Russell L., letter to Martin Gardner.
4. Beldon, T. H., "Runs of Squares," The Mathematical Gazette, Dec.
1961, 334-335.
5. Taussky, Olga, "The Many Aspects of the Pythagorean Triangles,"
Linear Algebra and Its Applications43: 285-295, 1982;
the following references appear in [5].
6. Edwards, H.M., "The Genesis of Ideal Theory," Arch. Hist. Exact
Sci.23: 321-378, 1980.
7. Friedrichs, K., From Pythagoras to Einstein, 1965.
8. Taussky, Olga, "Sums of Squares," Amer. Math. Monthly
77 : 805-830, 1970.