[EventHorizon]

"It is well known that there exist pairs of distinct squares that add to make another square.

For example, 5^2 + 12^2 = 25 + 144 = 169 = 13^2.

But this can be extended to any number of squares. For example, 2^2 + 5^2 + 14^2 = 15^2.

Prove that there exists a sum of n distinct squares that is also square."

[EventHorizon]

3^2 + 4^2 = 5^2...

and (3*x)^2 + (4*x)^2 = (5*x)^2 for any non-negative integer x

[EventHorizon]

First, multiplying two squares results in another square. It is easy to see this.... X^2 * Y^2 = X*X*Y*Y = (X*Y)^2.

Base cases:

If N=0, sum of nothing is 0 = 0^2.

If N=1, sum of any one square is the same square.

If N=2, any Pythagorean triple will work. I'll use 3^2 + 4^2 = 5^2.

Induction:

Assume you can find a sum of N-1 distinct squares whose sum is square.

Multiply all squares, except the smallest, by 5^2.

The smallest square will result in two new squares.

Multiply the smallest square by 4^2 and add a new square which is the old smallest square multiplied by 3^2.

Summing these two new squares made from the smallest square in the previous list results in 5^2 times the previous smallest square (e.g., X*3^2 + X*4^2 = X*(3^2+4^2) = X*5^2).

Each of the other squares were simply multiplied by 5^2.

This means that the sum will be 5^2 times the previous sum...which is another square.

This means the new list is N distinct squares whose sum is square.

Here's a quick example...in case my description was hard to understand.

Assume A^2 + B^2 + C^2 + .... + Y^2 is a list of N-1 distinct squares that sum's to Z^2.

Without loss of generality, assume A^2 is the smallest square in the list.

A^2 + B^2 + C^2 + .... + Y^2 = Z^2

5^2*(A^2 + B^2 + C^2 + .... + Y^2) = 5^2*Z^2

5^2*A^2 + 5^2*B^2 + 5^2*C^2 + .... + 5^2*Y^2 = 5^2*Z^2

3^2*A^2 + 4^2*A^2 + 5^2*B^2 + 5^2*C^2 + .... + 5^2*Y^2 = 5^2*Z^2

(3A)^2 + (4A)^2 + (5B)^2 + (5C)^2 + .... + (5Y)^2 = (5Z)^2

This final equation is a list of N distinct squares whose sum is also square.

[Guest]

"It is well known that there exist pairs of distinct squares that add to make another square.

For example, 5^2 + 12^2 = 25 + 144 = 169 = 13^2.

But this can be extended to any number of squares. For example, 2^2 + 5^2 + 14^2 = 15^2.

Prove that there exists a sum of n distinct squares that is also square."

I'm lost as to what you're asking for...

The main way I know these "square sets" is from Pythagoras, and there are infinite I believe, but I'm not understanding what you want to be proven?

[Guest]

I'm lost as to what you're asking for...

The main way I know these "square sets" is from Pythagoras, and there are infinite I believe, but I'm not understanding what you want to be proven?

He seems to be proving that if you have a set of N distinct squares whose sum is a square, you can achieve a set of N+1 distinct squares through an application of the Pythagorean theorem and some simple multiplication of the equation's members. It thus follows that since we have obvious and simple sets of 1,2, and 3 member squares which add up to another square, then we can achieve any set of N distinct squares which do the same. It's an interesting point, but I didn't really understand what he was getting at either until I looked at the answer.

[EventHorizon]

I'm lost as to what you're asking for...

The main way I know these "square sets" is from Pythagoras, and there are infinite I believe, but I'm not understanding what you want to be proven?

Pythagorean triples are two distinct squares that sum to another square.

This problem (I took from another website) is to prove that for any integer n, that you can find n distinct squares whose sum is square. (eg, prove that pythagorean quadruples, quintiples, and so on exist)