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,ca, b, c

b2=(a×c)+1\color{#3D99F6}\boxed{b^2 = (a \times c) + 1}

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

252=24×26+1    625=624+1    625=62525^2 = 24 \times 26 + 1 \implies 625 = 624 + 1 \implies 625 = 625


Relation 2 :

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

(a×b)+2c+a=c2\color{#D61F06}\boxed{(a \times b) + 2c + a = c^2}

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

(3×4)+2×5+3=52    12+10+3=25    25=25(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,ba, b

(a×b)a=a2\color{#CEBB00}\boxed{(a \times b) - a = a^2}

Example : let us take two numbers 6, 7

(6×7)6=426=36=62(6 \times 7) - 6 = 42 - 6 = 36 = 6^2


Relation 4 :

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

a2b2+2c=3(or)(a+b)(ab)+2c=3{\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)(78)+2(9)=15+18=3(7 + 8)(7 - 8) + 2(9) = -15 + 18 = 3


Relation 5 :

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

a2+2a+2b=b2(or)a2+4a+4=b2{\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

132+2(13)+2(15)=169+26+30=22513^2 + 2(13) + 2(15) = 169 + 26 + 30 = 225


Relation 6 :

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

a2b2=a+b(or)b2=a2(a+b)(or)a2=b2+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

312302=961900=61=31+3031^2 - 30^2 = 961 - 900 = 61 = 31 + 30


Relation 7 :

Sum of any nn consecutive integers is given by :

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

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

n = 5 (odd)

Median = 5+12=62=3th term =13\frac{5 + 1}{2} = \frac{6}{2} = 3^{\text{th}} \text{ term } = 13

Sum=median×n=13×5=65Sum = 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
1 year, 4 months ago

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Comments

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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

Mohmmad Farhan - 1 year ago

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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)

Mohmmad Farhan - 1 year ago

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I will explain the third relation.

a(a+1)=a2+aa^2+a

After you minus the a off you get a2a^2

Mohmmad Farhan - 1 year ago

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

Ram Mohith - 1 year ago

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

Mohmmad Farhan - 1 year ago

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The 6th relation works because

a2b2a^2-b^2 =(aba-b)(a+b)a+b)

Since they are consecutive, a-b=1 and thus 1(a+b)= (a2b2a^2-b^2)=(a+b)a+b)

Mohmmad Farhan - 1 year ago

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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)

Mohmmad Farhan - 1 year ago

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You may want to add something in the 4th4^{th}.

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

Mohmmad Farhan - 11 months, 3 weeks ago

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I think the 5th5^{th} is related to the binomial theorem

Mohmmad Farhan - 11 months, 3 weeks ago

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@Ram Mohith, For the 3rd3^{rd} relation you may not want to use yellow because it is hard to read it.

Mohmmad Farhan - 11 months, 3 weeks ago

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