# Interesting Relations Between Numbers

Hi Everyone,

I had been working out on numbers for quite some time and I had got some interesting relations among numbers which I will write below. There are from some small and common observations and it is not a big deal to prove them.

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Relation 1 :

For any three consecutive integers $a, b, c$

$\color{#3D99F6}\boxed{b^2 = (a \times c) + 1}$

Example : let us take the numbers 24, 25, 26

$25^2 = 24 \times 26 + 1 \implies 625 = 624 + 1 \implies 625 = 625$

Relation 2 :

For any three consecutive integers $a, b, c$

$\color{#D61F06}\boxed{(a \times b) + 2c + a = c^2}$

Example : let us take 3 numbers 3, 4, 5

$(3 \times 4) + 2 \times 5 + 3 = 5^2 \implies 12 + 10 + 3 = 25 \implies 25 = 25$

Relation 3 :

For any two consecutive integers $a, b$

$\color{#CEBB00}\boxed{(a \times b) - a = a^2}$

Example : let us take two numbers 6, 7

$(6 \times 7) - 6 = 42 - 6 = 36 = 6^2$

Relation 4 :

For any three consecutive numbers $a, b, c$

${\color{#EC7300}\boxed{a^2 - b^2 + 2c = 3}} \qquad \text{(or)} \qquad {\color{#EC7300}\boxed{(a + b)(a - b) + 2c = 3}}$

Example : let us take three numbers 7, 8, 9

$(7 + 8)(7 - 8) + 2(9) = -15 + 18 = 3$

Relation 5 :

For any two consecutive even or odd numbers $a, b$

${\color{#20A900}\boxed{a^2 + 2a + 2b = b^2}} \qquad \text{(or)} \qquad {\color{#20A900}\boxed{a^2 + 4a + 4 = b^2}}$

Example : let us take two consecutive odd numbers 13, 15

$13^2 + 2(13) + 2(15) = 169 + 26 + 30 = 225$

Relation 6 :

For any two consecutive numbers $a, b$ such that $a > b$

${\color{#624F41}\boxed{a^2 - b^2 = a + b}} \quad \text{(or)} \quad {\color{#624F41}\boxed{b^2 = a^2 - (a + b)}} \quad \text{(or)} \quad {\color{#BA33D6}\boxed{a^2 = b^2 + a + b}}$

Example : let us take two numbers 31, 30

$31^2 - 30^2 = 961 - 900 = 61 = 31 + 30$

Relation 7 :

Sum of any $n$ consecutive integers is given by :

$\color{#D61F06}\boxed{Sum = Median \times n}$

Example : find the sum of 11, 12, 13, 14, 15

n = 5 (odd)

Median = $\frac{5 + 1}{2} = \frac{6}{2} = 3^{\text{th}} \text{ term } = 13$

$Sum = median \times n = 13 \times 5 = 65$

That's all for now. Hope these relations will help you. If any mistakes are there please inform me .

If you too have any kind of relations with you please put it in form of a comment below and I will write them in this note in your name. Note by Ram Mohith
2 years, 5 months ago

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The first relation works because according to the formula. (a-1)(a+1)=a^2-1 (I also noticed this but you already put it up)

The third is because you have one extra group, namely a, and you just minus it off.

The seventh worked because you are essentially multiplying c on both sides. ( I also noticed this)

All the others are just WOW

- 2 years, 1 month ago

How about the relation for the Pythagorean triple (3,4,5). Multiplying the set by any whole number gives you another Pythagorean triple. Like multiply it by 2 you get (6,8,10)

- 2 years, 1 month ago

I will explain the third relation.

a(a+1)=$a^2+a$

After you minus the a off you get $a^2$

- 2 years, 1 month ago

Yes you are correct. In fact, you can prove all these relations by just taking the first term as $a$ second term as $a + 1$ and third term as $a + 3$.

- 2 years, 1 month ago

I am confused with a in English and Math in your comment

- 2 years, 1 month ago

The 6th relation works because

$a^2-b^2$ =($a-b$)($a+b)$

Since they are consecutive, a-b=1 and thus 1(a+b)= ($a^2-b^2$)=($a+b)$

- 2 years, 1 month ago

The eighth one works because you can distribute a 1 to 12 from 14 and you get 2 13's and distribute a 2 to 11 from 15 and you get 2 13's. It is kind of the Gauss Theorem of pairing numbers whose average is the same as the middle number (If there is)

- 2 years, 1 month ago

You may want to add something in the $4^{th}$.

Since a and b are consecutive instead of writing $(a-b)$ you can write $(-1)$

- 2 years ago

I think the $5^{th}$ is related to the binomial theorem

- 2 years ago

@Ram Mohith, For the $3^{rd}$ relation you may not want to use yellow because it is hard to read it.

- 2 years ago