My Math Forum p(x) for real x when p(p(x))+p(x)=x^4+3x^2+3

 Algebra Pre-Algebra and Basic Algebra Math Forum

 July 19th, 2017, 10:00 PM #11 Global Moderator   Joined: Dec 2006 Posts: 18,232 Thanks: 1437 You should find p(x) â‰¡ xÂ² + 1.
 July 20th, 2017, 12:46 AM #12 Newbie   Joined: Jul 2017 From: Denmark Posts: 11 Thanks: 0 Okey, how?
 July 20th, 2017, 12:54 AM #13 Global Moderator   Joined: Dec 2006 Posts: 18,232 Thanks: 1437 I assumed that p(x) is a polynomial in x, so it's a quadratic with no "bx" term (else p(p(x)) + p(x) would have a term in xÂ³). Can you make progress from there?
 July 20th, 2017, 12:37 PM #14 Senior Member   Joined: May 2016 From: USA Posts: 855 Thanks: 345 Skipjack "saw" how to get a unique solution without going through convolutions of algebra. But you can prove his answer is the unique solution if you are willing to do a bit of algebra. $\text {Given: }p(x) = ax^2 + bx + c \text { and } p(p(x)) + p(x) = x^4 + 3x^2 + 3.$ $\therefore p(p(x)) = p(ax^2 + bx + c) = a(ax^2 + bx + c)^2 + b(ax^2 + bx + c) + c =$ $a\{(ax^2 + bx)^2 + 2c(ax^2 + bx) + c^2\} + abx^2 + b^2x + bc + c =$ $a\{(a^2x^4 + 2abx^3 + b^2x^2) + 2acx^2 + 2bcx + c^2\} + abx^2 + b^2x + c(b + 1) =$ $a\{a^2x^4 + 2abx^3 + x^2(2ac + b^2) + 2bcx + c^2\} + abx^2 + b^2x + c(b + 1) =$ $a^3x^4 + 2a^2bx^3 + x^2(2a^2c + ab^2) + 2abcx + ac^2 + abx^2 + b^2x + c(b + 1) =$ $a^3x^4 + 2a^2bx^3 + x^2(2a^2c + ab^2 + ab) + x(2abc + b^2) + c(ac + b + 1).$ $\therefore p(p(x)) + p(x) = a^3x^4 + 2a^2bx^3 + x^2(2a^2c + ab^2 + ab + a) + x(2abc + b^2 + b) + c(ac + b + 2).$ Equating coefficients gives $a^3 = 1 \implies a = 1 \implies a^2 = 1 \text { and } ac = c.$ $2a^2b = 0 \implies 2 * 1 * b = 0 \implies b = 0.$ $c(ac + b + 2) = 3 \implies c^2 + 2c - 3 = 0 = (c - 1)(c + 3 )\implies c = 1 \text { or } c = -\ 3.$ $\therefore p(x) = x^2 + 1 \text { or } p(x) = x^2 - 3.$ Let's check. $p(x) = x^2 + 1 \implies p(p(x)) = (x^2 + 1)^2 + 1 = x^4 + 2x^2 + 2 \implies$ $p(p(x)) + p(x) = x^4 + 2x^2 + 2 + x^2 + 1 = x^4 + 3x^2 + 3.$ CHECKS $p(x) = x^2 - 3 \implies p(p(x)) = (x^2 - 3)^2 - 3 = x^4 - 6x^2 + 6 \implies$ $p(p(x)) + p(x) = x^4 - 6x^2 + 6 + x^2 - 3 = x^4 - 5x^2 + 3.$ NO. $\text {THUS } p(x) = x^2 + 1.$ I am not sure that skipjack's more intuitive and much easier method shows that his answer is unique. Last edited by JeffM1; July 20th, 2017 at 12:52 PM.
 July 20th, 2017, 06:23 PM #15 Global Moderator   Joined: Dec 2006 Posts: 18,232 Thanks: 1437 How did you know that $a$ is real?
July 20th, 2017, 07:01 PM   #16
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Joined: May 2016
From: USA

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Quote:
 Originally Posted by skipjack How did you know that $a$ is real?
Ahh. You are correct. My solution is not necessarily unique. Then you should have determined whether the other potential solutions were feasible.

I apologize to the OP for possibly misleading him and congratulate skipjack on winning the Houdini prize of the day.

 Tags 3x2, algebra, polynomails, ppx, pxx4, real

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