Fun With Sums

It’s been a while since there has been any math on the blog, so I figured I’d share a recent (trivial) mathematical fact I came upon while passing the time. A less noble goal is that I hope some of you will find it interesting enough to think about it for a while. In other words, I’m too lazy to keep working on it but I hope some others will fall into my trap and let me know the answer.

Several nights ago, a friend of mine was procrastinating on the Internet — something I hear people do from time to time. He chanced upon a random (nerdy) web comic where the author declared that

$10^2+11^2 + 12^2 = 13^2 + 14^2.$

I think basically everyone knows the Pythagorean triple

$3^2 + 4^2 = 5^2,$

but I had never come across (or wondered privately) whether the pattern continued. A small group of us got together to determine an answer. Indeed, one can find $m$ consecutive integers whose squares add up to the sum of the next $m-1$ squares. For example,

$21^2+22^2+23^2+24^2 = 25^2 + 26^2 + 27^2$

$36^2+37^2+38^2+39^2+40^2 = 41^2 + 42^2 + 43^2 + 44^2$

In general, the problem is to find $k$ which solves the equation

$\displaystyle{\sum_{i=0}^{m}(i+k)^2=\sum_{i=m+1}^{2m}(i+k)^2}$

Of course $k$ must depend on $m$ . Solving this equation is pretty simple using the following identities:

$\sum_{i=0}^m i=\frac{1}{2}m(m+1)$

$\sum_{i=0}^m i^2=\frac{1}{6}m(2m+1)(m+1)$

Using these identities correctly in our equation produces the polynomial

$k^2-m^2-2m^3-2km^2=0$

Plugging in $k=\alpha m$ produces

$m^2(m+1)(\alpha -2m-1)=0$

which forces the choice $\alpha=2m+1$ , i.e. $k=m(2m+1)$ . So, indeed, we can find a sequence of $2m+1$ consecutive integers so that the squares of the first $m+1$ of them add up to the sum of the squares of the last $m$ .

Now, it’s also true that

$3^3+4^3+5^3=6^3$

It is an exercise left to the reader to figure out the general pattern for cubes, if it exists, and then generalize. Homework is due next week.

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14 Responses to “Fun With Sums”

Carl Says:
February 23, 2009 at 9:40 am | Reply
x^4/4+x^3/2+x^2/4=(x(x+1)/2)^2
Aaron F. Says:
February 24, 2009 at 12:31 am | Reply
Oooooh, cool!
Jerrell Says:
March 3, 2009 at 2:44 am | Reply
Just dropping by.Btw, you website have great content!

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Qiaochu Yuan Says:
March 9, 2009 at 3:04 pm | Reply
Unfortunately, there’s no solution to i^3 + (i+1)^3 + (i+2)^3 + (i+3)^3 = (i+4)^3 + (i+5)^3 in integers; the polynomial is irreducible. This remains true if we add an extra term on the LHS. I think we should only expect finitely many such cubic identities; the resulting Diophantine in k and m probably has genus 1.
Luis Says:
March 11, 2009 at 6:26 am | Reply
No link to http://abstrusegoose.com/63 ?
Vineet Says:
March 12, 2009 at 12:43 pm | Reply
How do you ascertain that m|k ?
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March 18, 2009 at 12:57 pm | Reply
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Peter Says:
March 23, 2009 at 11:32 pm | Reply
That link to abstruse goose was actually what started this thing. I was reading the whole archive of it in an attempt to avoid work, saw that formula, and then succeeded in avoiding more work by working on it. The other Peter and I were talking about it for a bit, and afterwards we both came up with this answer.
Peter Luthy Says:
March 25, 2009 at 4:40 pm | Reply
Qiaochu, I was allowing that the pattern no longer had to be anything like the original. For instance, perhaps one cannot guarantee that all the terms will be consecutive — i.e. maybe the cubes on the left and right sides of the equation may be separated by some large gap. Also the number of terms on each side might not be linearly related any longer.

And thanks for the link Peter. I wasn’t sure what comic it was that started the whole discussion in the first place.
Peter Luthy Says:
March 25, 2009 at 6:55 pm | Reply
Vineet,

There are quite a few arguments for why we know that $k$ is what it is. I simply made a quick guess when I was working on the problem and it turned out to be correct. But here is an answer that might be more to your satisfaction: note that $k^2-m^2-2m^3-2km^2=0$ is a quadratic polynomial in $k$ where the coefficients are $a=1$ , $b=-2m^2$ , and $c=-m^2-2m^3$ . Applying the quadratic formula (and applying some algebra), one arrives with two solutions, the one derived in the post, and $k=-m$ which we don’t really want to consider anyway — it is silly because the terms squared in each sum will end up being symmetric around zero.
Vineet Says:
March 26, 2009 at 1:48 am | Reply
@Peter
Yeah, got it. k = m(m+2), and hence k=(alpha)m.
I actually wrote my query wrong, I meant m|k.

Thanks.
Academic Career Links Says:
March 31, 2009 at 10:47 pm | Reply
Nice post!
Let’s Read the Internet! week 10 « Arcsecond Says:
May 11, 2009 at 4:26 am | Reply
[...] Fun With Sums Peter Luthy at The Everything Seminar [...]
Bunny Doherty Says:
August 17, 2010 at 1:51 pm | Reply
Hi Peter,
HOW LONG HAS IT BEEN SINCE WE HAD OUR LAST COCKTAIL IN MANHATTAN? Well over 10 years I would guess. How have you been? hOPEFULLY YOU ARE THE PETER WHO WAS MY SALES REP MANY HAPPY YEARS AGO
BUNNY

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Fun With Sums

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