## November 2023

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

Find all whole numbersaandbthat 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 3*k*, *or* to 3*k* + 1, *or* to 3*k* + 2.

What would *b*² + 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) Dividingb² + 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).