Square + its Root + Next Root = Next Square; Product of 4 Consecutive Integers + 1 = Square

Square + its Root + Next Root = Next Square

We began to speak of the language of the “root” of a square. As an extension of exploring the difference of consecutive squares as a sequence of consecutive odd numbers, those odd numbers are a sum of two root numbers. The two root numbers are the root of the first square number and the root of the next root number. 

For example, if you have the square number 16 which is 4 squared (or 4^2), if you add the root 4 to 16 and then add the next root 5 (16 + 4 + 5), the sum is 25, the next square number (5^2). Let’s take the larger square number of 81=9^2; if you add 81 + 9 +10, you get 100=10^2. The algorithm can be expressed algebraically which I shared with the children: 

x^2 + x + (x + 1) = (x + 1)^2

Again, my intention is to expose the children to algebraic expression so when they get there in 6-8th grade, they are as comfortable as counting to 10 by 1s. My main objective was to ensure that the children mastered their vertical addition skills with carries.

Product of 4 Consecutive Integers + 1 = Square

The more advanced algorithm we explored this week with grades 3 and up required the children to multiply four consecutive integers and add one. The resulting number is always a square. Even more interesting is that the square number has a root which is the product of the first and fourth of the consecutive numbers plus one.

For example, 2x3x4x5+1=121 or 11^2, notice that if you multiply 2x5+1=11. For a larger example, 4x5x6x7+1=841=29^2; notice that if you multiply 4x7+1=29. My objective for this lesson was to have them discover a sequence in the square numbers as you climb from 1x2x3x4+1 to 2x3x4x5+1 to 3x4x5x6+1 to 4x5x6x7+1 and so on. The roots are 5, 11, 19, 29, 41, 55, and so on. This sequence has a difference of consecutive even numbers (6, 8, 10, 12 …). 

More importantly, I was looking for the children to use their powers of commutative property (1x2x3x4 = 2x4x1x3) where the order of how you multiply does not matter. For example, when multiplying 3x4x5x6, if you choose to multiply 3x5 and then 4x6 and then multiply their products, 15 x 24, you do get the product of 360. However, if you multiply 3x4=12 and then 5x6=30, your next job is easier 12 x 30 (12 x 3 = 36 and tac on a zero for 360). Maybe even better, if you multiply 3x6=18 and then 4x5=20, your next job is easier 18 x 20 (18 x 2 = 36 and tac on a zero for 360). I challenged the children to be strategic with their multiplication. When multiplying 6x7x8x9, I expected some of them to hit a wall since there are no multiples of zero. (54)(56) or (42)(72) is much more challenging. I introduced the standard algorithm for many of the older students and this is all about place value. For other students, the more reliable method of distributive property again allows students to use multiples of 10 and addition; thus 42(70 + 2) would yield 2940 + 84 = 3,024 or 56(50 + 4) would yield 2800 + 224 = 3,024.

We also were able to introduce an algorithm for multiplying by 11. For the 8x9x10x11 product, first 8x9=72, then 72x10=720 and then for 720x11 simply add 10 of “them” to 1 of “them.” So 7200 + 720 = 7,920. 

Next week, stay tuned for my proprietary algorithm for estimating square roots. These estimates are within .08 for the square root of 2 of the actual square root to within .0006 for the square root of 401. We will be using graphing calculators to check our work.