I bet you can't solve this puzzle!

You have a cube, which is made out of 10x10x10 smaller cubes. Can you use all the smaller cubes to make two other cubes?

If that doesn't work, you can also try using a different sized cube, not one with a side length of 10 cubes but any other number and see if you can come up with a solution for that.

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Northwest · M

Fermat proposed that this is impossible for anything larger than 2x2x2. Fermat passed before providing a proof.

Andrew Wiles proved it some 330 after Fermat's passing.

Luke73 · 22-25, M

@Northwest Yes Fermat‘s last conjecture.

DrWatson · 70-79, M

Here is another example. Frederick Mosteller, in the book "Fifty Challenging Problems in Probability", relates a problem posed by E.C. Molina which gives a concrete realization of the "Fermat Conjecture" (as it was known at the time) in terms of probability.

Two urns contain the same total numbers of balls, some black and some white in each. From each urn are drawn n >= 3 balls with replacement. Find the number of drawings and the composition of the two urns so that the probability that all white balls are drawn from the first urn is equal to the probability that the drawing from the second is either all whites or all blacks.

If we let

z = number of white balls in the first urn
y = number of white balls in the second urn
x = number of black balls in the second urn

For ease of typing on SW, let T = x + y. (the number of balls in each urn)

Then we need

( z / T )^n = ( x / T )^n + ( y / T )^n

Or

z^n = x^n +y^n

At the time that the book was published (1965) this was known to be impossible for n < 2000.

DrWatson · 70-79, M

This is equivalent to finding positive-integer solutions to the equation a^3 +b^3 = c^3 (in your example, c = 10). According to Wiles' theorem (formerly known as the Fermat conjecture) this is impossible.

But you have given a great example of how to visualize the content of that theorem concretely.

Luke73 · 22-25, M

@DrWatson Exactly

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