For a 6-year old:

- Make a filled in square. (She made a 3x3 square).
- How many pennies are there in the square? How many rows? How many columns? (She wanted to know which way do rows go?)
- Can you make a smaller square? (She made a 2x2 square). How many pennies? How many rows? How many columns?
- Can you make a smaller square? (She thought for a minute and put a single penny down. And laughed.) One row, one column, and one penny.
- Can you make an even smaller square? (She thought maybe not. I pointed at an empty space on the table. She laughed and said "zero rows, zero columns, zero pennies.

For the 11-year old.

- Make a sequence of triangles (the first one contains 1, the second 3, the third 6, the fourth 10). How much do you add to get from one to the next?
- If you take two triangles of base 4 (10 pennies each) and put them together what do you get (a rectangle that is 5 by 4).
- Can you see way to use that to calculate the number of pennies in a triangle of base size 10? (After some thinking he came up with that two of them were 10x11, so one of them would be 55.)
- Make a sequence of squares. How much do you add to get from one to the next. (You add 3 then 5 then 7 then 9...)
- So the sum of odd numbers from 1 to somewhere is a square.

For the 15-year olds: (They didn't actually touch the pennies. Nooo.. They can do it in their heads...)

- You remember what a pythogorean triple is? (Three integers
*a,b,c*such that*a*^{2}+*b*^{2}=*c*^{2}.) - How many are there? (Lots).
- Can you show there are an infinite number? (Well...)
- Consider a square of side length n and one of side length
*n*+1. What's the difference in their sizes? (2*n*+1). - Can 2
*n*+1 be square? (They now jumped ahead to the conclusion. Every odd number when squared is odd. So the first triple is 2*n*+1 = 1, so you get 0,1,1 (that's cheating!) The second is 2*n*+1=9, so you get 3,4,5. The third is when 2*n*+1=25 so you get 5,12,13. (Dana chimed in that you should have those memorized). The next one is 2*n*+1=49 so you get 7,24,25. - In general for any integer
*k*, let*l=*2*k*+1, then*l*^{2}is an odd square. The resulting pythogorean triple is*l,*(*l*^{2}-1)/2, and (*l*^{2}+1)/2.

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