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Arithmetic Sequences of Perfect Squares{1, 25, 49} is a 3-element arithmetic progression of perfect squares. Are there any 4- or more-element arithmetic progressions of perfect squares? Maybe we can answer this question by better understanding the 3-element sequences. 25-1 = 24, and 49-25 = 24. These are the differences of squares. So (5-1)(5+1) = 24, and (7-5)(7+5) = 24 From this we see there are three integers, a, b, and c such that (b-a)(b+a) = (c-b)(c+b) Here's an interesting pattern I notice when glancing at the thousands of 3-element sequences produced by my investigation: The constant differences are all multiples of 24. Why should that be? I imagine that if I explore all the various ways the constant differences can be expressed, I'll see why they must contain two factors of 2 and one factor of 3. So let's start looking at facts about the constant difference between these squares. If the arithmetic progression consists of {a, b, c} then it's true that b²-a² = c²-b². This also means that (b-a)(b+a) = (c-b)(c+b), and that equals (c-a)(c+a)/2. Do I see the factor of 24 here? No, frankly, I don't. So here are my questions:
The answer to the first question is given on the next page. The answer to the second question is much harder. Some effort to understand it is given on subsequent pages. Related pages in this website |
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The webmaster and author of the Math
Help site is Graeme McRae.
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