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A Neat Little Puzzle That Quickly Explodes💥

Friday 10 April 2026 at 21:54
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Edited by Richard Walker, Friday 10 April 2026 at 23:54

Here's a question I've seen here and there in various forms. This is my version

Double me, you get a square. Triple me you get a cube. I am the smallest such. What number am I?

Sometimes it's captions '90% of people can't solve this' or similar. But it's actually not too hard.

If the number we seek is n then we want ('double me') two times n to be a square and this can be achieved by making n twice a square, say n equals two times k squared . Then equation sequence part 1 two times n equals part 2 two multiplication two times k squared equals part 3 four times k squared equals part 4 left parenthesis two times k right parenthesis squared a square as required.

But we also want ('triple me') equation sequence part 1 three times n equals part 2 three multiplication two times k squared equals part 3 six times k squared to be a cube and the smallest value of k that makes this work is k equals six , and equation sequence part 1 n equals part 2 two multiplication six squared equals part 3 two multiplication 36 equals part 4 72 .

And sure enough, equation sequence part 1 two multiplication 72 equals part 2 144 equals part 3 12 squared and equation sequence part 1 three multiplication 72 equals part 2 216 equals part 3 six cubed , a square and a cube exactly as we want.

By why stop there? Can the idea be extended? We can't extend the pattern to four (I'll put a proof in the Comments tomorrow) but we can make it work with five . To see how to do this let's look at the prime factors of our previous example, 72 equals two cubed multiplication three squared .

When we multiply this by two we get equation sequence part 1 two multiplication two cubed multiplication three squared equals part 2 two super three plus one multiplication three squared equals part 3 two super four multiplication three squared and both exponents are even so this is a square.

When we multiply it by three we get equation sequence part 1 three multiplication two cubed multiplication three squared equals part 2 two cubed multiplication three super two plus one equals part 3 two cubed multiplication three cubed and both exponents are multiples of three so this is a cube.

Using this idea but now with two , three and five (after a fair bit of working) we find the exponents work in the way we want if we take

n equals two super 15 multiplication three super 20 multiplication five super 24

You can see that if we multiply this by two all the exponents will be divisible by two , if by three they all divisible by three , and if by five divisible by five , just as we want. This is the smallest number that meets our goal but doesn't look small, here it is all 31 digits, a big jump from the 2 digits of 72 !

6810125783203125000000000000000

We can continue in this way as long as we like: adding seven gives 233 digits

150462810922326152710290228433686961530697356776074449373600141938371053848189980134027578261857302770024765419887333164323078738017254430529707573248000000000000000000000000000000000000000000000000000000000000000000000000000000000000

and the numbers just continue growing in a super-exponential way. I suppose there must something we could say about the long term behaviour but it's beyond my technical capabilities.

Still I might be able to write up a description and get it into the Online Encyclopedia of Integer Sequences (OEIS). I'll give it a try.

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Richard Walker
Saturday 11 April 2026 at 15:37
by Richard Walker

As promised

If in whole numbers two times n was a square p squared and four times n a fourth power q super four , then we would have

equation sequence part 1 q super four divided by p squared equals part 2 two times n divided by n equals part 3 two

which implies left parenthesis q squared divided by p right parenthesis squared equals two and so we would have Square root of two equals q squared divided by p , a rational number. But this is impossible because Square root of two is irrational.