square numbers

• May 2nd 2010, 06:48 AM
coolminty
square numbers
Hey!

I am trying to solve this problem...

2^2+3^2+6^2=7^2

and

3^2+4^2+12^2=13^2

(^2 represents squared)

I have found out that the first two numbers need to be consecutive. The third number is the product of the first two and the 'answer' is the sum of the third number plus 1. This works for all numbers but i am not sure why!?!?!?

Any suggestions?
• May 2nd 2010, 07:10 AM
Tikoloshe
Well, can you write an expression for each term in terms of an algebraic variable? For example, call the first term $n^2$, the second term $(n+1)^2$, and so on. Then simplify the equation to a tautology (something which is obviously true) such as $n=n$.
• May 2nd 2010, 07:21 AM
coolminty
thanks-

i have done the left hand side of the equation as followed by i want to know why such is the case!

Why does this happen? What are the underlying mathematical reasons?

x^2 + (x+1)^2 + (x(x+1))^2
• May 2nd 2010, 08:11 AM
HallsofIvy
Quote:

Originally Posted by coolminty
Hey!

I am trying to solve this problem...

2^2+3^2+6^2=7^2

and

3^2+4^2+12^2=13^2

(^2 represents squared)

I have found out that the first two numbers need to be consecutive. The third number is the product of the first two and the 'answer' is the sum of the third number plus 1. This works for all numbers but i am not sure why!?!?!?

Any suggestions?

What problem are you trying to solve? Certainly NOT that
"2^2+3^2+6^2=7^2" and "3^2+4^2+12^2=13^2" because those are trivial arithmetic: 4+ 9+ 35= 13+ 36= 49 and 9+ 16+ 144= 25+ 144= 169.

Are you trying to find some general patterns so that a sum of three squares is equal to a square?
• May 2nd 2010, 08:42 AM
coolminty
Thank you for your message hallsofivy!

I have established that:-

The first two numbers in the equation are consecutive (eg $2^2$ and $3^2$)
The third number is the product of the first two (eg $6^2$
The 'answer' is one more that the third number (eg $7^2$

What i want to know is why is this the case? What is special about using these rules and numbers that makes the rules work.
• May 2nd 2010, 09:20 AM
Tikoloshe
Quote:

Originally Posted by coolminty
i have done the left hand side of the equation

Good. Now “do” the RHS of the equation (what is one more the than the third number, quantity squared?) and manipulate the resulting equation until you arrive at a tautology (FYI upon working out your example, I arrived at $2x(x+1)=2x(x+1)$, although there are many other ways). As long as the steps you used are reversible, you have an answer as to why this is pattern holds for arbitrary numbers.
• May 2nd 2010, 09:25 AM
chiph588@
Quote:

Originally Posted by coolminty
thanks-

i have done the left hand side of the equation as followed by i want to know why such is the case!

Why does this happen? What are the underlying mathematical reasons?

x^2 + (x+1)^2 + (x(x+1))^2

$x^2 + (x+1)^2 + (x(x+1))^2 = (x^2+x+1)^2$
• May 3rd 2010, 03:19 AM
simplependulum
To obtain this identity we can put $a = x , b = x+1$

we have $a^2 + b^2 + a^2 b^2$

$= (a^2 - 2ab + b^2 ) + a^2 b^2 +2ab = (a-b)^2 + 2ab + a^2b^2 = 1 + 2ab + (ab)^2 = (ab+1)^2$

$= (x^2 + x + 1)^2$
• May 4th 2010, 09:05 AM
hollywood
Quote:

Originally Posted by coolminty
Thank you for your message hallsofivy!

I have established that:-

The first two numbers in the equation are consecutive (eg $2^2$ and $3^2$)
The third number is the product of the first two (eg $6^2$
The 'answer' is one more that the third number (eg $7^2$

What i want to know is why is this the case? What is special about using these rules and numbers that makes the rules work.

Every odd number is the difference of two consecutive squares:

$(n+1)^2-n^2=(n^2+2n+1)-n^2=2n+1$

so since the first two numbers are consecutive, one is even and the other is odd, so the sum of squares is odd.

$x^2+(x+1)^2=2m+1$, where $m=\frac{x^2+(x+1)^2-1}{2}=\frac{x^2+(x^2+2x+1)-1}{2}=\frac{2x^2+2x}{2}=x^2+x$.

So

$x^2+(x+1)^2=2m+1=(m+1)^2-m^2$ = $(x^2+x+1)^2-(x^2+x)^2=(x^2+x+1)^2-(x(x+1))^2$

which gives

$x^2+(x+1)^2+(x(x+1))^2=(x^2+x+1)^2$, the exact equation given by chiph588@.