»  Solutions to puzzles in my VDARE.com monthly Diary

  November 2023

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In the Math Corner of my November 2023 Diary I posted the following brainteaser.

Find all whole numbers a and b that satisfy the equation

          a (a + 1) (a + 2) = b² + 4.

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•  Solution

I'll label that equation (i). So

(i)   a (a + 1) (a + 2) = b² + 4.

Which whole numbers a and b satisfy (i)?

The solution is in the remainders:  precisely, in the fact that

(ii)  If you divide any whole number by 3, the remainder is either 0, or 1, or 2.

Starting with the left-hand side of (i), notice that a, (a + 1), and (a + 2) are three consecutive numbers. It follows that precisely one of them must divide exactly by 3, leaving remainder zero.

So we are multiplying three numbers, one of which divides exactly by 3. The result of that multiplication must itself divide exactly by 3. So

(iii)  The left-hand side of (i) leaves remainder 0 when divided by 3.

What about the right-hand side of (i)? First, it follows from (ii) that any whole number must, for some other whole number k, be equal either to 3k, or to 3k + 1, or to 3k + 2.

What would b² + 4 be in those three cases? It would be either

9k² + 4

or

9k² + 6k + 5

or

9k² + 12k + 8

Dividing each of those by 3 you will get remainders 1, 2, and 2 respectively. So

(iv)  Dividing b² + 4 by 3 will never yield remainder 0.

From results (iii) and (iv) it follows that (i) can never be true. There are no whole numbers a and b that satisfy (i).