# Sum of Cubes Equals Square of Sum

By Stephen Meskin

A standard application of mathematical induction is the proof that for any positive integer n:

13 + 23 + … + n3 = (1 + 2 + … + n)2               (1)

It is a striking result. You might have seen the proof in school; I will not reproduce it here, but you may want to try to prove it.

The interesting thing about (1) is that it is not the only “sum of cubes equal to square of sum.” For any n > 1, there are at least two different sets of integers 1 ≤ a1a2 ≤ … ≤ an such that:

a13 + a23 + … + an3 = (a1 + a2 + … + an)2         (2)

For example, take n = 2. Then 2, 2 works as well as 1, 2. Moreover, these are the only two solutions for n = 2, because if b ≥ max(a, 3) then

a3 + b3a2 + (b – 1)b2 + b2a2 + 2ab + b2 = (a + b)2

Observe that if a > 1, the first inequality is strict and if a = 1, the second inequality is strict.

Problem 1. Show that there are exactly two solutions to (2) for n = 3 (That is, exhibit the two solutions and show that there are no others.)

Problem 2. Show that, in addition to (1), there is a second general solution to (2) for all n.

Problem 3. Find all the solutions to (2) for n = 4.

By the way for n = 6, the number of solutions is 18.

Solutions may be emailed to cont.puzzles@gmail.com.

In order to make the solver list, your solutions must be received by Sept. 30, 2019.

Solutions to Previous Puzzle—Celebrating Retirement If a watermelon fell the same distance over the first 3 seconds compared to the last 3 seconds of its descent, how far did the watermelon drop? We could use our math skills to solve these problems, knowing that velocity is the first derivative of distance and that acceleration is the first derivative of velocity, but it is cleaner and more fun to solve via the tried-and-true 2-dimensional motion physics equations that we all learned in high school. To figure out how far the watermelon dropped in the first 3 seconds, use the physics equation for change in distance = vo × t+ 0.5 × at2. From the problem we know v0 = 0, t = 3, and a = 10. Solving, we see the watermelon fell 45 meters in the first 3 seconds (and during the final second). We also know due to the constant force of gravity that the velocity at impact is 10 meters per second squared less than the velocity a second beforehand. Hence, we can use the equation that distance = t × (v + v0)/2 over the final second of the drop, where the distance is 45 meters, t = 1, and v = v0 + 10. Solving for v, we see that v = 50. As the watermelon had an initial velocity of zero and increased its speed 10 meters per second every second, we now know the watermelon had a total flight time of five seconds, and dropped 0.5 × 10 × 52 meters, or 125 meters.

How far did Big T’s watermelon travel horizontally after leaving the roof with a velocity of 5 meters per second? First, we need to know how long the watermelon was in the air before impact. To solve this, we will first analyze the problem only looking at the vertical direction of the watermelon, and use the equation distance = vo × t + 0.5 × at2. The problem stated that the initial velocity in the vertical direction = 0, distance =10, and a = 10. Solving, t = (2)0.5. As the horizontal velocity of the watermelon is 5 and does not change, we can now easily see that the horizontal distance traveled is simply 5 × (2)0.5, or very close to 7.07 meters.

Solvers

Steve Alpert, Robert Bartholomew, Robert Bender, Scott Boulay, Geoff Bridges, Bob Byrne, William Carroll, Bob Conger, Deb Edwards, David Engelmayer, Bill Feldman, Chris Fievoli, Mark Fowler, Jerry Francis, Yan Fridman, Mark Glickman, Olivier Guilbaud, Rui Guo, Clive Keatinge, Ken Kudrak, Lincoln Financial Group Team, Michael Murrell, Jim Muza, Paul Navratil, Scott Parker, David Promislow, Joe Schmuecker, Noam Segal, Lenny Shteyman, Brian Six, John Snyder, Mark Spong, Al Spooner, Ronald Stokes, Doug Szper, Daniel Wade.

 Corrections to the solution to “Four Points,” July/August 2019: 1. Solver David Promislow should have had an asterisk next to his name for providing a full solution. 2. The vertical line of the rhombus should be light blue (because it is short). Thanks to the avid readers of Contingencies for reporting these corrections.
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