POTW Returns!

Snow? Bah! Problem Of the Week doesn't care about snow.

That's right, it's time for another semester of teasers, enigmas, and conundrums. Our theme this term is:


NUMBER THEORY, WITH MATH JOKES

As always, we start with a fairly easy problem. Overall, though, I think the problems this term are a little more challenging than my usual fare, so don't get cocky!

Feel free to post solutions in the comments. Though this is nominally a competition, I'm not worried that my students are going to read this blog, desperately looking for the solution to the problem. This is entirely voluntary for the students, and the kind who choose to participate are also not the kind who are going to try to cheat.

More like this

Why
won’t Goldilocks drink from a glass with exactly
eight pieces of ice in it? Because it’s too
cubed

A variation: Why are cans of Boston Baked Beans restricted to have at most 239 beans?

One more and they'd be toofawty.

If you look at a^3+(a+1)^3+(a^2)^3 = (a+3)^3, expand it out and simplify, a lot of things fall out and you end up with a(a^2-6) = 9.

If a is a positive integer, then that means it (and a^2-6) have to be divisors of 9, which give 1, 3, and 9 as the only possibilities. a=1 gives a(a^2-6) = -5, which doesn't work. a=9 gives a(a^2-6)= 675, which doesn't work, but a=3 gives a(a^2-6) = 9, which is what we want.

By Buddha Buck (not verified) on 26 Jan 2016 #permalink

dean--

A classic! I love it.

Now for a quick test to see if Latex is supported in the comments.

$latex x^3+4x^2-2x+3=0$

Hope this works!

Success!

Pay no attention to that equation in the previous comment. Your TeX code should be surrounded by “$latex” on the left and “$” on the right. There is only limited TeX capability, though, so don't get too fancy. Normal math symbols are fine, but I don't think matrices or lining up equations works.

I used the form K^3 = 6K+9 (where K, K+1, etc. are the 4 consecutive integers). 3 divides the RHS so it must divide the LHS, so K=3m, which reduces to 3m^3 = 2m +1.

The LHS mod(m) is zero and the RHS mod(m) is 1, which only works for m=1. So K is uniquely 3(1)= 3, and the four integers are 3,4,5, and 6.

Yes, this one is straightforward algebra. I started with $latex (n-1)^3 + n^3 + (n+1)^3 = (n+2)^3$, which simplifies to $latex n^3 - 3n^2 - 3n - 4 = 0$. The cubic expression factors into $latex (n - 4) (n^2 + n + 1)$. The linear factor gives the solution as 3, 4, 5, and 6. The quadratic factor has a pair of complex conjugate roots, so this is the only solution involving integers (including negative integers).

By Eric Lund (not verified) on 26 Jan 2016 #permalink

Same as the above posters, I just did some basic algebra and ended up with 3, 4, 5, 6. Since it's a cubic equation, and cubic equations only ever have at most one solution, this is the only possible answer.

I did it the same way as Buddha Buck and Jim V: use the lowest number in the sequence as my unknown, expand, get x^3 = 6x+9, solve from there.
Its interesting that Eric Lund used the second one as the unknown. The square terms don't drop out that way. Eric what motivated you to set the second term as your unknown?

eric@8: I didn't immediately notice that treating the smallest integer as the unknown would result in the quadratic terms canceling on both sides; choosing the second integer as the unknown causes the quadratic terms on the left side to cancel.

A side effect of my method is that the roots of the quadratic factor I obtained are the complex cube roots of 1. Add 1 to each and you get the complex cube roots of -1. The other two numbers turn out to be the values of $latex \sqrt{3}(\mp i)^{1/3}$ that are off the imaginary axis, so of course the cube of the one with the largest real part is the sum of the cubes of the other three (the middle two cancel, and the ones on the ends are equal).

By Eric Lund (not verified) on 27 Jan 2016 #permalink

Well, this hardly qualifies as a math joke, but it's as good as the ice cubes.

What's the best time to go to the dentist?

tooth hurty (2:30)

Although writing down the cubic and spotting its linear factor is clearly the Right Answer, it's worth noting that you can also observe that (1) for large numbers, the sum of three consecutive cubes is clearly always going to be bigger, and in fact (2) it seems like the numbers shouldn't need to be very large for that to happen, and therefore (3) if there's a solution it probably involves small numbers -- and, so motivated, just try a few small examples starting at 1,2,3 and working upward.

I think this actually finds the solution faster than doing the algebra does, for me, but of course it depends on how quick you are with algebraic manipulation and with arithmetic.

(The algebraic approach lets you cleanly eliminate the possibility of other solutions, which is a substantial advantage. You can formalize the idea that "for larger n, the sum of three cubes is too big" but I think it ends up more work than just solving the whole thing algebraically.)

GAZZA,

Surely you meant something else by your comment that cubic equations have at most one solution. What if your cubic equation had reduced to x^3-6x^2+11x-6 = 0, for example? That equation (as you can easily check) has solutions x=1, x=2 and x=3. In general cubic equations have 3 solutions, although two may be complex conjugate pairs and there may be multiple roots (such as x^3+3x^2+3x+1 = 0 for example, which has x=-1 as a three-fold root since the left side is equal to (x+1)^3).