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

Let a = (v² + u²)/2,  b = u*v,  c = (v² - u²)/2, then

a*(1 + x²) + 2bx + c*(1 - x²) = (1 + x²)(v² + u²)/2 + 2uvx + (1-x²)(v² - u²)/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 + x²) + 2*30*x - 8*(1 - x²) = 25x² + 30x + 9 =  (5x + 3)²

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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 :/

Alan, is your answer the same as mine ?  I think it is :/

Yes.  My b² + c² equals my a².  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.

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. :)

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