Sums of Squares Generating Other Sums of Squares
Here’s an interesting relationship. We’ll take 3 single digit squares which when summed, equal 3 other single digit numbers squared and summed:
42 + 52 + 62 = 82 + 32 + 22
We’ll use these 6 digits to generate 3 two digit numbers. We use one number from the left side and pair it with another number from the right to create our two digit number. (And it doesn’t matter which number from the left we pair with the right.) Now, if we take the sum of the square of each of these numbers that we created, we find that it is equal to the sum of each of those numbers reversed and squared. Let’s consider some examples. Using the relationship mentioned above, we’ll match the first number on the left with the first number on the right, the second with the second and the third with the third to give us the numbers 48, 53 and 62. Now notice that the sum of the square of each of these numbers is equal to the sum of the same numbers reversed and squared (48 reversed is 84, 53 reversed is 35 and 62 reversed is 26):
482 + 532 + 622 = 842 + 352 + 262
But since it doesn’t matter which number from the left is paired with which number from the right, these other sets can also be generated from the same 6 numbers:
482 + 522 + 632 = 842 + 252 + 362
432 + 582 + 622 = 342 + 852 + 262
432 + 522 + 682 = 342 + 252 + 862
422 + 582 + 632 = 242 + 852 + 362
422 + 532 + 682 = 242 + 352 + 862
In each case, we are generating the numbers on the left from our 6 single digit numbers. The numbers on the right are the reverse of the numbers on the left. As remarkable as this is, it is by far not the only example. It can also be done with these two sets of four numbers.
82 + 52 + 32 + 22 = 72 + 62 + 42 + 12
Matching a number on the left with a number on the right gives us:
872 + 562 + 342 + 212 = 122 + 432 + 652 + 782
or also:
862 + 542 + 312 + 272 = 272 + 132 + 452 + 682
842 + 512 + 372 + 262 = 622 + 732 + 152 + 482
812 + 572 + 362 + 242 = 422 + 632 + 752 + 182
Altogether, there are 20 more possibilities, making a total of 24 different possibilities.
And, it doesn’t even matter if there are 0's in the relationship, it is still true. For example, we’ll take a Pythagorean triple and add a zero to the hypotenuse side of the relationship, thus giving us two numbers on each side of the equality:
32 + 42 = 52 + 02
Which will generate these equalities:
352 + 402 = 042 + 532
302 + 452 = 542 + 032
Or we could also begin with this generator:
12 + 32 + 52 + 92 = 42 + 62 + 82 + 02
The possibilities here are amazing. However, let’s extend this idea just a little further. Instead of using a sum of single digits squared, let’s try it with a set of two digit numbers squared. We’ll treat each two digit number as a single entity and pair a left side entity with a right side entity. The sum of each of these terms squared should be equal to the sum of the left-right pair reversed and squared. Let’s look at an example:
122 + 562 + 642 = 242 + 322 + 762
From this example, we can create the following relationships, and sure enough, they work:
12242 + 56322 + 64762 = 24122 + 32562 + 76642
12322 + 56762 + 64242 = 32122 + 76562 + 24642
12762 + 56242 + 64322 = 76122 + 24562 + 32642 ..., etc.
Again, all we have done is take a number on the left and match it with a number on the right. Pairing all of the numbers in this way creates the terms for the left side of the equality. To create the numbers on the right side, we reversed each of the pairs on the left.
Knowing this, we could use any Pythagorean Triple and the number zero to create a valid pair of numbers. However, we must keep in mind that if at least one of the numbers being used is a two digit number, then all of the numbers must be two digit numbers. For example:
52 + 122 = 132 + 02
Actually becomes:
052 + 122 = 132 + 002
Which will generate the following equalities:
05132 + 12002 = 13052 + 00122
05002 + 12132 = 13122 + 00052
At first glance this relationship seems really amazing, and it definitely is something you can use to impress your friends. But how do we know that it always works? Even though we looked at several variations of this formula, we will content ourselves with just proving the basic relationship. The other variations are extensions of this basic example.
We start with: a2 + b2 = c2 + d2
Where a, b, c and d are all single digit numbers. And given that this first statement is true, we want to show that the following is also always true:
(a*10 + c)2 + (b*10 + d)2 = (c*10 + a)2 + (d*10 + b)2
First, we expand the squares on each side to get:
100a2 + 20ac + c2 + 100b2 + 20bd + d2 = 100c2 + 20ac + a2 + 100d2 + 20bd + b2
Next, we combine similar terms, and we have:
99a2 + 99b2 = 99c2 + 99d2
We divide the relationship by 99, and it reduces back to our original equation:
a2 + b2 = c2 + d2
Palindromic Sums of Squares
Let’s return to an earlier relationship where we used 32 + 42 = 52 + 02 to create the two sets of equalities
352 + 402 = 042 + 532 and 302 + 452