**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 (2×1) + 1. Look at the difference between 16 and 25, or 42 and 52 respectively. 9 = (2×4) + 1.

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

Qn = n2

and

Qn+1 = (n + 1)2.

The difference between Qn and Qn+1 is given as:-

Qn+1 – Qn = (n + 1)2 – n2

= (n + 1)(n + 1) – n2

= n2 + 2n + 1 – n2

**Qn+1 – Qn = 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 Rn and Rn+1 where

Rn = n3

and

Rn+1 = (n + 1)3

R = Rn+1 – Rn

= (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*.

if we know square of number n then we can compute square of number m by below formula

m^2 = n^2 +(m-n) +(m+n)

special case : m=n+1

(n+1)^2 = (n^2) +(1)*(2n+1)

Similarly differences between adjacent integers of higher powers can be computed by using binomial theorem and symmetric property of binomial coefficient.

e.g. for cube of number binomial coefficient are

1 3 | 3 1

by observation of above coefficient it is clear that coefficient are symmetric about center (or | line).This property can be used to combine two term and simplify formula to compute differences between adjacent integers of higher powers.

for odd n formula to compute differences between adjacent integers of higher powers

(x+1)^n – x^n =summation(1, (n-1)/2 )[c(n,k)* ( x^(n-k)+ x^k ) ] + 1

Similarly we can derive formula to compute differences between adjacent integers of higher powers n which is even.

Please replace the first formula m^2 = n^2 +(m-n) +(m+n) with m^2 = n^2 +(m-n)*(m+n)

Hi Alex,

Do you know this method to calculate the cube of a number ?

Suppose that n^3 and (n-k)^3 are given.

Then (n+k)^3 = 2*n^3 + 6 * n * k^2 – (n-k)^3

( it's the result of A^3 + B^3)

example : 16^3 = (10+6)^3 = 2*10^3 + 6*10*6^2 – (10-6)^3 = 4096

Evidently you have to do a good choice for the first number and for n

And it's sometimes a goog idea to do the calculus in 2 or 3 parts.

So, if you want to obtain the value of 896,

Choice 16 as starting number and n=60 => 60-44=16 and 60+44=104

then you obtain easily 104^3

104^3=2*60^3+6*60*44^2-16^3=1124864 (for 16^3 see above)

Now choose n=500 : 104=500-396, thus 500+396=896 and the method gives

896^3=2*500^3+6*500*396^2-104^3=719323136

Best regards

N.