Solve this math prob for a gmail invite

Discus: College Confidential Café: Solve this math prob for a gmail invite
 By Lisasimpson (Lisasimpson) on Saturday, October 02, 2004 - 04:01 pm: Edit

find a,b for the parabola y = ax^2 + bx + 3 if the vertex is (2,4)

i have 4 invites left to give, so the first person to solve it can have how many they want and then anybody who verifies the answer can have one too

 By Aspirer42 (Aspirer42) on Saturday, October 02, 2004 - 04:08 pm: Edit

For another Gmail invite, prove that

(x^n) + (y^n) = (z^n) has no solutions for n > 2, when x, y, and z are non-zero integers.

Show all work.

 By Xasuke (Xasuke) on Saturday, October 02, 2004 - 05:40 pm: Edit

I'll do the parabola one.. I don't want an invite.

y = ax^2 + bx + 3

plugging in the given vertex (2, 4):
4 = 4a + 2b +3

ok.. and using x=-b/2a for the x cord. of the vertex:
b=-4a

substitute:
4=4a -8a +3
a=-1/4
so..
b=1

finally:
y = -.25x^2 + x + 3

Right now, I'm too lazy to think about the other problem.. Pretty much prove the Pythagorean theorem?

 By Lisasimpson (Lisasimpson) on Saturday, October 02, 2004 - 05:53 pm: Edit

gracias xasuke

could that be a trick prob aspirer? cuz of x y and x are all 1 then the problem will work no matter what n is..unless i'm just reading it wrong or something.

 By Cherrybarry (Cherrybarry) on Saturday, October 02, 2004 - 07:26 pm: Edit

where's andrew wiles when you need him?!!!

 By Thermodude (Thermodude) on Saturday, October 02, 2004 - 07:27 pm: Edit

well...posting a 187 page proof on CC would take a while anyways....too bad Fermat just couldn't have written down his proof somewhere...

 By Ubercollegeman (Ubercollegeman) on Sunday, October 03, 2004 - 01:06 am: Edit

I'd solve it, Aspirer, but the proof is too long for the contents of this message box to contain...