What does this equation mean to you?

[Willowz]

Hi first time on this forum. Ok I was bored and found a way to calculate numbers to the power of 2:

a b c d e f $$\infty$$
1 2 3 4 5 6 $$\infty$$

b-1=a c-2=a d-3=a e-4=a f-5=a g-6=a

$$b^{2}=a^{2}+a+b \vee b^{2}=a^{2}+2a+1$$
$$c^{2}=a^{2}+2(2a+2) \vee c^{2}=a^{2}+4b$$
Same for the rest d,e,f,g,...$$\infty$$

What am I getting at? Well I'm at my second year at high scool , and well I still don't have the "software" :-) to find any other mening of this. I don't really know what I made up exept counting powers in my head using what's above.

Thanks for any input!

 

[CRGreathouse]

Hi first time on this forum. Ok I was bored and found a way to calculate numbers to the power of 2

I don't understand your method. How would it find 2^8, for example?

 

[Willowz]

I don't understand your method. How would it find 2^8, for example?

What I meant is that you can raise any number with this equation but only to the power of ^2

 

[ramsey2879]

Hi first time on this forum. Ok I was bored and found a way to calculate numbers to the power of 2:

a b c d e f $$\infty$$
1 2 3 4 5 6 $$\infty$$

b-1=a c-2=a d-3=a e-4=a f-5=a g-6=a

$$b^{2}=a^{2}+a+b \vee b^{2}=a^{2}+2a+1$$



$$c^{2}=a^{2}+2(2a+2) \vee c^{2}=a^{2}+4b$$



Same for the rest d,e,f,g,...$$\infty$$

What am I getting at? Well I'm at my second year at high scool , and well I still don't have the "software" :-) to find any other mening of this. I don't really know what I made up exept counting powers in my head using what's above.

Thanks for any input!

What you are doing is finding a relation between two numbers and their second powers (squares of the numbers), not powers of 2 which means 2 muptiplied by itself a certain number of times, 2,4,8,16,32, ... In algebra, there is a way to multiplied b^2 = (a+1)*(a+1) to get the square of b since b = a + 1. You got it right when you got the result a^2 + 2a + 1. Similaly c^2 = (a+2)*(a+2). You wrote c^2 = a^2 + 4b which is correct given your values for a,b and c. But a^2 + 4b = c^2 can Be true for whole values of a,b and c only if $$c = a + 2m$$ where m is also a whole number (or integer). Try it for different values of a and c. It is correct also if c = a+ 4 and b = 2a + 4. Try it for various values of a and c. You would be making a good start in number theory if you could show that c^2 could equal a^2 + 4b where a, b and c are each integers, if c = a + n where n = 2m (or an even number) but not if n = 2m + 1 (or an odd number).

 

[Willowz]

Thanks ramsey2879 for the info, again sorry for my bad math lingo.

 

[CRGreathouse]

What I meant is that you can raise any number with this equation but only to the power of ^2

Ah, I see. You can square any number. So can you show me how you would square 12 with this method?

 

[Willowz]

Ah, I see. You can square any number. So can you show me how you would square 12 with this method?

:
a=11 b=12
b^2=121+11+12 or b^2=121+22+1
144=144 144=144

a=10 b=11 c=12
c^2=100+2(20+2) or c^2=100+4(11)
c^2=100+44 c^2=100+44
144=144 144=144

 

[ramsey2879]

:
a=11 b=12
b^2=121+11+12 or b^2=121+22+1
144=144 144=144

a=10 b=11 c=12
c^2=100+2(20+2) or c^2=100+4(11)
c^2=100+44 c^2=100+44
144=144 144=144

This is simple to show via algebra:

$$c = a + 2$$
$$c^2 = (a+2)*(a+2)$$
$$= a^2 + 4a + 4 = a^2 + 4(a + 1)$$
$$= a^2 + 4b$$

or if c = a + 2n then
$$c^2 = (a + 2n)(a+2n)$$
$$= a^2 + 4na + 4n^2$$
$$= a^2 + 4b \quad \| b = n(a+n)$$