**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 1

^{2}, 2

^{2}, 3

^{2}, 4

^{2}and 5

^{2}.

Here are the first five square numbers, and for completion let's throw in 0

^{2}as well:-

Number | Square | Difference |
---|---|---|

0^{2} | 0 | N/A |

1^{2} | 1 | 1 |

2^{2} | 4 | 3 |

3^{2} | 9 | 5 |

4^{2} | 16 | 7 |

5^{2} | 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, 1

^{2}and 2

^{2}. The difference, 3, can be considered (2x1) + 1. Look at the difference between 16 and 25, or 4

^{2}and 5

^{2}respectively. 9 = (2x4) + 1.

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

_{n}= n

^{2}

and

_{n+1}= (n + 1)

^{2}.

The difference between Q

_{n}and Q

_{n+1}is given as:-

_{n+1}- Q

_{n}= (n + 1)

^{2}- n

^{2}

= (n + 1)(n + 1) - n

^{2}

= n

^{2}+ 2n + 1 - n

^{2}

**Q**.

_{n+1}- Q_{n}= 2n + 1Let's take this formula to look at the difference between two adjacent squares for an arbitrary value, say 37

^{2}. What is the next square number, 38

^{2}, going to be?

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

A quick calculation, and I work out that 37

^{2}= 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.

^{2}= 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 n

^{2}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, 999

^{2}(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, 4

^{3}and 5

^{3}, or between 1741

^{3}and 1742

^{3}, or between n

^{3}and (n + 1)

^{3}?

Let's see.

Number | Square | Difference |
---|---|---|

0^{3} | 0 | N/A |

1^{3} | 1 | 1 |

2^{3} | 8 | 7 |

3^{3} | 27 | 19 |

4^{3} | 64 | 37 |

5^{3} | 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

_{n}= n

^{3}

and

_{n+1}= (n + 1)

^{3}

R = R

_{n+1}- R

_{n}

= (n + 1)

^{3}- n

^{3}

= (n + 1)(n + 1)(n + 1) - n

^{3}

= n

^{3}+ 3n

^{2}+ 3n + 1 - n

^{3}

= 3n

^{2}+ 3n + 1

= 3n(n + 1) + 1.

**R = 3n(n + 1) + 1.**

Can we confirm that this formula works? Let's try three examples, namely 5

^{3}and 6

^{3}, 74

^{3}and 75

^{3}, and 1741

^{3}and 1742

^{3}.

^{3}= 125.

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

R = (15 x 6) + 1

R = 91

**125 + R = 216 = 6**.

^{3}Next, let's look at 74

^{3}and 75

^{3};

^{3}= 405224

R = (3 x 74 x 75) + 1

R = (222 x 75) + 1

R = 16651

125

^{3}+ R = 405224 + 16651

= 421875

**= 75**

^{3}.And now the last one, just to make sure:-

^{3}= 5,277,112,021

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

= (5223 x 1742) + 1

= 9098466 + 1

= 9098467

**5277112021 + 9098467 = 5286210488 = 1742**.

^{3}Pretty conclusive, then.

**Conclusion**

So, then, it's pretty clear that to determine the neighbouring number for any given square number N = n

^{2}, you can use the formula

**2n + 1**

and to determine the neighbour of any cube number N = n

^{3}, 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*.

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