**10**and

^{2}+ 11^{2}+ 12^{2}= 365**13**

^{2}+ 14^{2}= 365So 365 is the sum of two sets of *consecutive squares*, but, more importantly, those two sets are themselves consecutive: {10, 11, 12} and {13, 14}.

Now that's interesting! Okay, so it's also a co-incidence, really. Sums of consecutive squares have to add up to *something*, and, occasionally, those "somethings" will be the same number. But can we write a general rule for this?

Of course, we can. If we can write it, we can (hopefully) solve it. (Again, *of course* I can solve it, or I wouldn't be asking the question, but I'm sure that there are many, many rules which we could pose which I, personally, couldn't solve. However, this is a simple one to work with.)

At the very simplest level, we have the sum of the squares of two consecutive positive integers. It's obvious that these two numbers can't equal the sum of two higher integers, so the sum has to be of only *one* number. Not much of a sum, I grant you, but we're starting with a trivial case.

We want to find consecutive positive integers, **a**, **b**, and **c** such that

**a**

^{2}+ b^{2}= c^{2}so we'll use the variable

*n*to stand in for the lowest integer,

*(n+1)*for the next consecutive integer, and

*(n+2)*for the third consecutive integer. Now we can rewrite the equation as

**n**

^{2}+ (n+1)^{2}= (n+2)^{2}Squaring the binomials, we get:

**n**

^{2}+ n^{2}+ 2n + 1 = n^{2}+ 4n + 4Combing like terms gives us:

**2n**

^{2}+ 2n + 1 = n^{2}+ 4n + 4Rewrite this as a quadratic equation by subtracting the right side of the equation from both sides:

**n**

^{2}- 2n - 3 = 0Which factors into: **(n - 3)(n + 1) = 0**. Therefore, **n = 3** or **n = -1**, but because we want a positive whole number, we'll discard the -1 and accept the 3. That makes the three consecutive integers **3, 4, 5** and, therefore, **3 ^{2} + 4^{2} = 5^{2} **.

But, of course, you already knew that. So why do all that work? Because now we can move up to four or five consecutive numbers. We can use the same procedure to find solutions to

**a**or

^{2}+ b^{2}+ c^{2}= d^{2}**a**

^{2}+ b^{2}+ c^{2}= d^{2}+ e^{2}To save space, and to be as annoying as those textbook writers of my youth, I'll leave *a, b, c, d* to you to try. I'll give you a hint: there aren't any positive integer solutions, but you can prove that for yourself instead of taking my word for it. Go ahead -- challenge authority!

For the sum of the squares of three consecutive positive integers equal to the sum of the squares of the next two integers, this is the equation we write:

**n**

^{2}+ (n+1)^{2}+ (n+2)^{2}= (n+3)^{2}+ (n+4)^{2}When the dust settles, what will be the value of *n*? If you didn't get it, you weren't paying attention. It's in the first paragraph of this article. The solutions are **n = 10** and **n = -2**. Once again, we toss the negative and we're left with: **{10, 11, 12}** and **{13, 14}**.

This brings two closing questions: The obvious question is what seven consecutive positive integers *a, b, c, d, e, f, g* fall into this pattern? (I didn't say that the *answer* was obvious, but it's easy to figure out.)

And another question about another pattern: In the first case, we threw out the solution **n = -1**. In the second case, we discarded ** n = -2**. I'll go ahead and tell you that in finding the answer to the next sequence of numbers, you'll have to get rid of the solution **n = -3**. My question: will the negative solution we discard always have the same absolute value as the number of terms on the right side of the equation?

I'll leave that as an exercise to the reader. You have a whole, exciting, brand New Year to work it out!

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