#2**+20 **

Let a = (v^{2} + u^{2})/2, b = u*v, c = (v^{2} - u^{2})/2, then

a*(1 + x^{2}) + 2bx + c*(1 - x^{2}) = (1 + x^{2})(v^{2} + u^{2})/2 + 2uvx + (1-x^{2})(v^{2} - u^{2})/2 = (ux + v)^{2}

Choose whatever numbers you like for u and v.

For example, if u = 3 and v = 5

a = 17, b = 15, c = -8 and 17*(1 + x^{2}) + 2*30*x - 8*(1 - x^{2}) = 25x^{2} + 30x + 9 = (5x + 3)^{2}

.

Alan
Aug 16, 2015

#1**+15 **

If a(1+x^2)+2bx+c(1-x^2) is a perfect square, what is a,b,c?

mmm I suppose

$$\\a(1+x^2)+2bx+c(1-x^2) \\\\

=a+ax^2+2bx+c-cx^2 \\\\

=ax^2-cx^2+2bx+a+c \\\\

=(a-c)x^2+2bx+(a+c) \\\\

\sqrt{(a-c)(a+c)}=b\\\\

a^2-c^2=b^2\\\\

a^2=b^2+c^2$$

Not sure if I can go further :/

Melody
Aug 16, 2015

#2**+20 **

Best Answer

Let a = (v^{2} + u^{2})/2, b = u*v, c = (v^{2} - u^{2})/2, then

a*(1 + x^{2}) + 2bx + c*(1 - x^{2}) = (1 + x^{2})(v^{2} + u^{2})/2 + 2uvx + (1-x^{2})(v^{2} - u^{2})/2 = (ux + v)^{2}

Choose whatever numbers you like for u and v.

For example, if u = 3 and v = 5

a = 17, b = 15, c = -8 and 17*(1 + x^{2}) + 2*30*x - 8*(1 - x^{2}) = 25x^{2} + 30x + 9 = (5x + 3)^{2}

.

Alan
Aug 16, 2015

#4**+5 **

Yes. My b^{2} + c^{2} equals my a^{2}. It's just that if you use your form, then you have to be more careful about the choice of b and c if you want a to be rational.

Alan
Aug 16, 2015

#5**+5 **

I don't care if a is rational. I am not rational so why should a be. That would not be fair. :))

Should b or not b be rational. Only c would know I suppose :)

Thanks Alan. :)

Melody
Aug 16, 2015

#6**+10 **

a(1+x^2)+2bx+c(1-x^2) =

a + ax^2 + 2bx + c - cx^2 =

(a - c)x^2 + 2bx + (a + c) = (√(a - c)x + √(a + c) ) ( √(a - c)x + √(a + c))

This implies that 2√(a^2 - c^2)x = 2bx → √(a^2 - c^2) = b

So, one solution is that any "Pythagorean Triple" for a, b and c would work [However, we might - or might not - have "rational" coefficients ]

For example, let a = 5, b= 4 and c = 3

Then we have:

(a - c)x^2 + 2bx + (a + c) =

(5 - 3)x^2 + 2bx + (5 + 3) =

2x^2 + 8x + 8 =

(√2x + √8) (√2x + √8) = (√2x + √8)^2

CPhill
Aug 16, 2015