## TheSmartOne one year ago Challenge Question: Factor $$\sf a^2 + b^2=?$$ Note: This isn't the Pythagorean Theorem The other similar formulas to that one, if you need it, are: $$\sf a^2-b^2 = (a+b)(a-b)$$ $$\sf a^2+2ab+b^2 = (a+b)^2$$ $$\sf a^2 - 2ab+b^2 = (a-b)^2$$ Hint: Think outside the box ;)

1. anonymous

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

2. anonymous

am I not supposed to post it? do I message or something?

3. TheSmartOne

well seems like it wasn't really hard at all, great job xD

4. UsukiDoll

$a^2 + b^2=?$ $$\sf a^2-b^2 = (a+b)(a-b)$$ $$\sf a^2+2ab+b^2 = (a+b)^2$$ (I'm taking this one) $$\sf a^2 - 2ab+b^2 = (a-b)^2$$ Thick outside the box? LIke this? $$\sf a^2+2ab+b^2 = (a+b)^2$$ $$\sf (a+b)(a+b) = (a+b)^2$$ [ using factoring] $$\sf (a+b)^2 = (a+b)^2$$ [same terms, so rewrite ] I tried T_T

5. anonymous

this isn't rewriting but $$a^2+b^2=\frac12[(a-b)^2+(a+b)^2]$$

6. anonymous

err this isn't factoring but it is rewriting

7. TheSmartOne

^

8. UsukiDoll

x( so it's like a proof... sort of.. T_T

9. TheSmartOne

Sort of... This equation blew my mind away when my Prof told us about it xD

10. UsukiDoll

it's like.. what I'm trying to apply is that for those problems... start at the left to achieve the right or vice versa. . . .

11. anonymous

(a*a)+(b*b)? @TheSmartOne

12. TheSmartOne