# Math Help - square numbers

1. ## 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?

2. 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 $n=n$.

3. 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

4. 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?

5. 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.

6. 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.

7. 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
$x^2 + (x+1)^2 + (x(x+1))^2 = (x^2+x+1)^2$

8. To obtain this identity we can put $a = x , b = x+1$

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

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

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

9. 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@.