Progression of Square and Cube Numbers

Crossposted to the Vedic Mathematics India blog:-

Square Numbers

Some of you may have studied the progression of square numbers, and discovered the rather elegant pattern which connects them - namely, that between two adjacent square numbers, the difference is always going to be an odd number; and with each step to the next square number, the difference always increases. This has always been a useful guide in predicting when each next square number should be; but what is the formula by which you can work out what that next square number is?

Well, let's look at the smallest square numbers - the square numbers for 1², 2², 3², 4² and 5².

Here are the first five square numbers, and for completion let's throw in 0² as well:-

Number	Square	Difference
02	0	N/A
12	1	1
22	4	3
32	9	5
42	16	7
52	25	9

As you can see, not only is the difference an odd number each time; it is a unique odd number. Other than 1 and 4, no other pair of square numbers will ever be found that will differ by 3; beside 25 and 36, no other pairing of square numbers will ever have a difference of 11.

The very specific progression between square numbers reveals a pattern, which can be formulated. But what is the formula for this difference between the squares?

Let's look at a pair of square numbers; 1 and 4, 1² and 2². The difference, 3, can be considered (2x1) + 1. Look at the difference between 16 and 25, or 4² and 5² respectively. 9 = (2x4) + 1.

Now let's abstract this to the case of the number n, where

Q_n = n²
and

Q_n+1 = (n + 1)².
The difference between Q_n and Q_n+1 is given as:-

Q_n+1 - Q_n = (n + 1)² - n²

= (n + 1)(n + 1) - n²

= n² + 2n + 1 - n²

Q_n+1 - Q_n = 2n + 1.
Let's take this formula to look at the difference between two adjacent squares for an arbitrary value, say 37². What is the next square number, 38², going to be?

Using the formula 2n + 1, and substituting n = 37, we arrive at:-

(2 x 37) + 1 = 74 + 1 = 75.
A quick calculation, and I work out that 37² = 1369. Adding 75 to that, we get 1444 - and the square root of 1444 is ... 38.

One last test, just in case; two adjacent large numbers, 13365 and 13366.

13365² = 178,623,225

(2 x 13365) + 1 = 26731

178623225 + 26731 = 178649956

sqrt(178649956) = 13366.
So there you have it. If you know one square number n² and its corresponding square root n, you can work out its adjacent neighbour by calculating the difference, 2n + 1.

But there's more. If you know the number n, you can work out the neighbour to n + 1, n + 2, n + 3 ... and also go back, subtracting 2(n-1) + 1, 2(n-2) + 1, 2(n-3) + 1 ..., always using the same formula, each time. Handy to know, say, if you were challenged to calculate not only, say, 999² (998,001 of course), but also the next ten square numbers above and below it ...

Cube Numbers

Does a similar relationship exist between adjacent cube numbers - between, say, 4³ and 5³, or between 1741³ and 1742³, or between n³ and (n + 1)³?

Let's see.

Number	Square	Difference
03	0	N/A
13	1	1
23	8	7
33	27	19
43	64	37
53	125	61

This pattern is not so easy to spot, so let's look at the difference R between R_n and R_n+1 where

R_n = n³
and

R_n+1 = (n + 1)³

R = R_n+1 - R_n

= (n + 1)³ - n³

= (n + 1)(n + 1)(n + 1) - n³

= n³ + 3n² + 3n + 1 - n³

= 3n² + 3n + 1

= 3n(n + 1) + 1.

R = 3n(n + 1) + 1.
Can we confirm that this formula works? Let's try three examples, namely 5³ and 6³, 74³ and 75³, and 1741³ and 1742³.

5³ = 125.

R = 3 x 5 (5 + 1) + 1

R = (15 x 6) + 1

R = 91

125 + R = 216 = 6³.
Next, let's look at 74³ and 75³;

74³ = 405224

R = (3 x 74 x 75) + 1

R = (222 x 75) + 1

R = 16651

125³ + R = 405224 + 16651

= 421875

= 75³.
And now the last one, just to make sure:-

1741³ = 5,277,112,021

R = ((1741 x 3) x 1742) + 1

= (5223 x 1742) + 1

= 9098466 + 1

= 9098467

5277112021 + 9098467 = 5286210488 = 1742³.
Pretty conclusive, then.

Conclusion

So, then, it's pretty clear that to determine the neighbouring number for any given square number N = n², you can use the formula

2n + 1
and to determine the neighbour of any cube number N = n³, you can use the formula

3n(n + 1) +1.
I leave the determination of the formulae to calculate the differences between adjacent integers of higher powers to the reader. One hint I will leave you with: look to Pascal's Triangle.