Problems

More on Pillow talks. Problem 3: 6/1/21

• Prove that the sum of two different squares, multiplied by the sum of two different squares, gives the sum of two squares in two different ways.

The sum of two different squares (x² +y²) multiplied by the sum of two different squares (m² +n²) is (x² +y²)(m² +n²)

This seems much harder than the first two problems.

I started by generating some data.

x = 1, y = 2 and m = 3, n = 4 would give 125. Which can be written as 2² + 11² or 5² + 10²

I saw the 11 first: 11 is 2 x 4 + 1 x 3 (that is yn + xm).

That means that I can start to fill in some parts of the equivalence I’m seeking.

(x² +y²)(m² +n²) = (yn + xm)² + ?² (the sum of two squares)

That is x²m² + x²n² + y²m² + y²n² = y²n² +2xymn + x²m² + ?²

Which means that ?² = x²n² - 2xymn + y²m² = (xn - ym)²

So, the sum of two different squares (x² +y²) multiplied by the sum of two different squares (m² +n²) can be written as (yn + xm)² + (xn - ym)² (the sum of two squares). This can be confirmed by expanding the brackets.

Let's test this.

Sum of two different squares multiplied by the sum of two different squares:

(4+16)(1+4) [ ie x = 2, y = 4, m=1 and n =2 ] should turn out to be (4x2 + 2x1)² + (2x2 - 4x1)²

So 20 x 5 should be 100 + 0. Which is true!

This is one solution. The original problem states that there is another solution too.

Going back to the original data:

x = 1, y = 2 and m = 3, n = 4 would give 125. Which can be written as 2² + 11² or 5² + 10²

2² + 11² is found from the expression above. So let’s look at 5² + 10².

10 is 1x4 + 2x3, ie xn + ym.

The rest is fairly easy. 5 is 2x4 - 1x3, ie yn - xm

So the other expression will be (xn + ym)² + (yn - xm)², which again can be confirmed by expanding.

Square Numbers & Pillow Talks: 30/11/20

3 problems from Lewis Carroll's Pillow Problems. In The Wine number 26 and The Wine number 27 I published 3 problems from Lewis Carroll's book Pillow Problems related to square numbers. These were:

• Is it always true that if you double the sum of two squares you get the sum of two squares? If so can you prove it?

• Prove that 3 times the sum of three squares is also the sum of 4 squares.

• Prove that the sum of two different squares, multiplied by the sum of two different squares, gives the sum of two squares in two different ways.

I always like to start these sorts of problems by looking at some numbers, to get a feel for what I'm trying to prove.

Problem 1: Double the sum of two squares is the sum of two squares

The Wine 26 actually gave some examples for this one.

2(5² + 3²) =2 (25 + 9) = 68 = 64 + 4 = 8² + 2²

2(7² + 4²) = 2(49 + 16) = 130 = 121 + 9 =11² +3²

It isn't too hard to spot a pattern here: In the first one 8 = 5 + 3 and 2 = 5 - 3. In the second one 11 = 7 + 4 and 3 = 7 - 4

So we might theorize that 2(x² + y²) = (x + y)² + (x - y)² . Expanding the brackets on the right hand side of that equation confirms that this is true.

Problem 2: 3 times the sum of three squares is also the sum of 4 squares

This is harder. One of the things which makes it harder is that it is quickly apparent that there are more combinations of numbers to try.

For example: 3(1²+2²+3²) = 42 = 1 + 9 + 16 + 16 or 0 + 1 + 16 + 16 or 1 + 1 + 4 + 36

Also 3(1²+3²+4²) = 78 = 1 + 4 + 9 + 64 or 0 + 4 + 25 + 49

Assuming that the numbers on the right hand side are sums or differences of the numbers on the left hand side (as in Problem 1), we can probably ignore the combinations with 0 in (as none of the three numbers on the left are the same). This leaves:

3(1²+2²+3²) = 42 = 1 + 9 + 16 + 16 or 1 + 1 + 4 + 36

3(1²+3²+4²) = 78 = 1 + 4 + 9 + 64

Look at that last one: as it happens 64 = (1 + 3 + 4)²

Looking back at 3(1²+2²+3²), we might notice that (1 + 2 + 3 )² = 36. IF this is the case, then we are now only looking at:

3(1²+2²+3²) = 42 = 1 + 1 + 4 + 36 and

3(1²+3²+4²) = 78 = 1 + 4 + 9 + 64

Then it isn't a big step to spotting that maybe

3( x² + y² + z²) = (y-x)² + (z-y)² + (z-x)² + (x + y + z)²

So for example 3( 1² + 3² + 5²) = (3-1)² + (5-3)² + (5-1)² + (1 + 3+ 5)² = 2² + 2² + 4² + 9² which is true, and the algebra can be verified by expanding.

I'm still working on Problem 3!