Expressing Certain Square Numbers as the Sum of Three Squares

Anyone who cannot cope with mathematics is not fully human. At best he is a tolerable subhuman who has learned to wear shoes, bathe, and not make messes in the house.

Robert A. HEINLEIN

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There are certain perfect square numbers that can be written as the sum of three perfect squares. In this blog post let us consider a few of these square numbers.

To begin with, consider the number 9. We can write it as

9 = 1 + 4 + 4                                        or, 32 = 12 + 22 + 22.                          [1]

This is an example of a single-digit square number (9 here). There are other square numbers also, which can be written as above. Consider the example of a two-digit square number.

49 = 4 + 9 + 36                                   or, 72 = 22 + 32 + 62;                          [2]

Few examples from three-digit square numbers are:

169 = 9 + 16 + 144                           or, 132 = 32 + 42 + 122;                     [3]

441 = 16 + 25 + 400                         or, 212 = 42 + 52 + 202;                     [4]

961 = 25 + 36 + 900                         or, 312 = 52 + 62 + 302;                     [5]

Consider few examples from four-digit perfect square numbers:

1849 = 36 + 49 + 1764                    or, 432 = 62 + 72 + 422;

3249 = 49 + 64 + 3136                    or, 572 = 72 + 82 + 562;

5329 = 64 + 81 + 5184                    or, 732 = 82 + 92 + 722;

8281 = 81 + 100 + 8100                  or, 912 = 92 + 102 + 902.

From the above examples, can we see any symmetry? Consider [2], for example. The product of 2 and 3 on the right is equal to 6. In [3], the product of 3 and 4 on the right is equal to 12. Similarly, in [4] the product of 4 and 5 on the right is equal to 20. In view of this symmetry, consider the following expression:

n2 + (n + 1)2 + {n(n + 1)}2

= n2 + (n2 + 2n + 1) + {n(n + 1)}2

= 1 + 2n2 + 2n + {n(n + 1)}2

= 1 + 2n(n + 1) + {n(n + 1)}2

= {1 + n(n + 1)}2.

From this identity, for various positive values of n, we can obtain various relations.

Again, if we add n2 in identity [2], we get the following relation:

9n2 = 1n2 + 4n2 + 4n2                                        or, (3n)2 = (1n)2 + (2n)2 + (2n)2                     [6]

Using this identity, we can find a few relations given below:

62 = 22 + 42 + 42;                92 = 32 + 62 + 62;                152 = 52 + 102 + 102 etc.

Similarly, from [3] we get the following identity:

49n2 = 4n2 + 9n2 + 36n2                                   or, (7n)2 = (2n)2 + (3n)2 + (6n)2;                   [7]

Your suggestions are eagerly and respectfully welcome! See you soon with a new mathematics blog that you and I call Math1089 – Mathematics for All!“.

2 comments

  1. We are familiar with Pythagorean triangles – integral solutions of equation a^2+b^2=c^2 ,where a,b ,c are natural numbers ;The problem is to find three dimensional a rectangular parallelepiped whose edges & Diagonal should be positive integers A^2+B^2+C^2=D^2
    If a=3=A.b=4=B,c=5;C=12;;D=13;
    Then 5^2+12^=13^2 Or 3^2+4^2++12^2=13^2 means parallelepiped whose edges are 3,4 &12 & Diagonal is
    13
    if n=3 ; A= n=3;B=n+1=4 ; C=n*(n+1)=3*4=12

    Liked by 1 person

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