Institute for Quantum Information and Matter, a National Science Foundation Physics Frontiers Center

# One line proof

It is not often that we come across a problem whose solution can fit in the margins of a notebook. In fact, many of the problems I have worked on in the field of quantum many-body physics require proofs that often exceed 30 pages. And that is without taking into account the several references included as sources for results used as “elementary” tools in the proof (referees love these papers…) So it is natural to think that a proof is a proof, so long as it is correct, and once confirmed by the academic community it is time to move on to something new.

But… sometimes things get interesting. A 30-page proof collapses to a 3-page proof when a different point of view is adopted (see the famous Prime Number Theorem). Below, you will find two problems that may, or may not have a “one-line” proof. The challenge is for you to find the shortest, most elegant proof for each problem:

A unit triangle: A triangle with sides \$latex ale b le c\$ has area 1. Show that \$latex b^2 ge 2\$.

Fermat’s Lost Theorem: Show that \$latex (x+y)^n = x^m+y^m\$ has only one solution with positive integers \$latex x > y > 0\$ and \$latex m,n > 1\$.

In case you figure out the above problems (who knows, you guys are pretty good), you can try your luck with this one:

Magic Square Cubed: Show that there is no \$latex 3times3\$ array of distinct cubed numbers (e.g. \$latex 8,27,125,ldots\$) such that the sums of numbers in each row, each column and each of the two diagonals are all equal. That is, show that there is no magic square of cubes.

And if you are an alien supergenius, you can go for all the money:

Magic Square Superpower Challenge: Show that there is no \$latex 3times3\$ array of n-th powers of distinct numbers with \$latex nge 4\$, such that the sums of numbers in each row, each column and each of the two diagonals are all equal. That is, show that there is no magic square of n-th powers, for \$latex n ge 4\$.

The most elegant solution to the Superpower Challenge will receive a \$100 prize. And in case you can show that there is no magic square of distinct squares… Nah, that’s impossible.

P.S.: To claim the prize, you and I should be able to understand your solution.

2017-01-13T10:06:02+00:00 October 19th, 2012|The expert's corner|19 Comments

1. blk October 19, 2012 at 7:19 am - Reply

A unit triangle: In a triangle a,b,c there is at most one angle >=90°. Let x,y be the two other angles <=90°, thus sin(alpha)<=1 and sin(beta)<=1. These two angles will be attached to the longest side c. Given two angles alpha, beta and the one side b (opposite of beta), we know the area A of a triangle is A=b² sin(alpha)sin(alpha+beta)/[2 sin(beta)]=1. Thus b²=2 sin(beta)/[sin(alpha)(sin(alpha+beta))]. But now sin(alpha)sin(alpha+beta)<=sin(alpha+beta)<=sin(beta), considering alpha,beta=1, therefore also b²>=2.

• blk October 19, 2012 at 7:20 am - Reply

After “Let x,y”: x=alpha, y=beta

• blk October 19, 2012 at 7:23 am - Reply

“…considering sin(alpha)<=1, sin(beta)<=1" in the last line.

2. Alex October 19, 2012 at 7:43 am - Reply

A unit triangle: A triangle with sides a <= b = 2.

Suppose b<2, then Area = b*h/2 = b*a*sin(theta)/2 < 2*sin(theta) <= 1.

• Alex October 19, 2012 at 8:02 am - Reply

Let me try writing that again…

Suppose b²<2, then Area = b*h/2 = b*a*sin(theta)/2 < 2*sin(theta)/2 <= 1.

• Anthony October 19, 2012 at 2:48 pm - Reply

Indeed, if b is the base, then the height is at most a, which less than b. The area is therefore at most b^2/2.

3. jasperavisser October 19, 2012 at 2:50 pm - Reply

Unit triangle:

Take b is baseline, h is height of triangle.
hb = 2
h <= a = 2

• jasperavisser October 19, 2012 at 2:52 pm - Reply

WordPress ate the last statement:
b² <= 2
b² <= 2

• jasperavisser October 19, 2012 at 2:53 pm - Reply

Ok, last try, now fully html-entitied:

Take b is baseline, h is height of triangle.
hb = 2
h <= a <= b
b² <= 2

4. spiros October 19, 2012 at 3:23 pm - Reply

The Unit Triangle answers above are the shortest, most elegant ones I can possibly imagine – one line proofs! Now on to Fermat’s Lost Theorem. Any geniuses willing to take a shot at it?

• Lubos Motl October 19, 2012 at 11:43 pm - Reply

\$latex (2+1)^2 = 2^3 + 1^3\$
because both sides are equal to nine. Unfortunately there’s not enough room on this line to prove that there are no other solutions. 🙂

5. Lubos Motl October 20, 2012 at 1:23 am - Reply

Magic square cubed.

If \$latex S\$ is the sum – of any row, column, or diagonal – then the sum of all 4 sums of lines going through the middle is \$latex 4S\$ but it is also equal to \$latex 4z^3+sum_{i=1}^8 x_i^3\$ where \$latex z^3\$ is the entry at the central square and \$latex x_i^3\$ are the remaining eight entries. The sum of all entries is also equal to \$latex 3S = z^3+sum_{i=1}^8 x_i^3\$. The difference implies \$latex S = 3z^3\$; the central entry is always the “average” of the entries. Then \$latex x^3+y^3+z^3=S\$ implies that \$latex x^3+y^3=2z^3\$ where \$latex x^3,y^3\$ are the opposing entries on a line going through the middle.

I read that \$latex x^3+y^3=2z^3\$ has no solutions with distinct positive integers, as proved by Euler and Legendre, but I am not able to reproduce the proof now although it shouldn’t be too hard.

The paid challenge has simply third powers replaced by higher powers in the text above. I think it is an extremely hard job for \$100 as the sources indicate that one needs to prove the higher-exponent counterpart of the previous paragraph but the known ways to do so rely on Wiles’ proof of Fermat’s Last Theorem. 😉

• spiros October 20, 2012 at 1:26 am - Reply

Indeed. Which is why I am wondering if there are unknown ways to prove it using the extra constraints from the square.

• Lubos Motl October 20, 2012 at 1:42 am - Reply

It’s conceivable but I think I may rather safely add \$100 to your bounty for that problem – for a solution not using Wiles’ results. 😉

• spiros October 20, 2012 at 5:12 pm - Reply

You are most likely right. Still, we are young. We can dream 🙂

6. Hiccup (@A_Hiccup) October 24, 2012 at 4:15 am - Reply

I am going to take a long shot here with the question regarding Fermat.

Let’s take an ellipse here:
(x/a)^2 + (y/b)^2 = 1 would define an ellipse.
Let x/a = g, y/b = h
Thus, (g + h)^m = g^2 + h^2 = 1 and that is the only possible solution with m,n> 1 and g> h> 0

If g and h are ellipse’s major and minor axis respectively..

I might be a bit confused here, too.

• spiros October 25, 2012 at 12:14 am - Reply

Dear Hiccup (cool name), I am about to post a partial solution to Fermat’s Lost Theorem. I hope you find it interesting and take the time to figure out the remaining part of the proof! Your questions will be answered within the upcoming post.

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8. James Tomson October 28, 2012 at 2:47 pm - Reply

Wile’s Proof of Fermat’s Last Theorem http://www.coolissues.com/mathematics/Wile'sproofofFLT.html