Giving the Pythagorean Theorem,

**a**, find three consecutive numbers for a, b and c.

^{2}+ b^{2}= c^{2}The answer, of course, is everyone's favorite Pythagorean Triple, 3, 4 and 5.

But suppose instead of three consecutive numbers, we had five consecutive numbers, which were split with the three smaller values on the left and the two greater values on the right? That would give us:

**a**

^{2}+ b^{2}+ c^{2}= d^{2}+ e^{2}The answer is in the previous post. I'll omit it here in case you want to work it out.

Now, a couple approaches could work here. The first, and probably best if you plan on going further, is to replace the variables with n, n+1, n+2, etc., and then using FOIL (or a "FOIL"-free alternative if you hate "FOIL"), combining like terms and solving the resulting equation.

The other, which I can use on Day 1, is guess and check. Okay, stop laughing and rolling your eyes, and hear me out.

First of all, most of my students haven't handled a scientific calculator very much let alone a TI-83, 84, or N-Spire. (Yes, we had a bunch donated to the school as part of a technology initiative. Unfortunately, they aren't allowed to be used during the Regents exams, so we have to switch back to the older calculators. But that's a rant for another day.) An activity like this could be a simple and thoughtful first exercise.

Second, many of them have little or no Number Sense or Estimating skills. How would they approach the problem? Would they try 3, 4, 5, 6, 7 first? When that doesn't work, will they move to 4, 5, 6, 7, 8, or will they jump a little higher a little faster? But if they don't try every combination, how will they know if they went too far and passed the answer?

Then, after finding the answer and comparing it to 3, 4, 5, what would their first guess be for

**a**?

^{2}+ b^{2}+ c^{2}+ d^{2}= e^{2}+ f^{2}+ g^{2}Naturally, all of this occurs to me during the last week of classes, after the

final exam has been given.

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