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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 
Senior Member Joined: May 2016 From: USA Posts: 855 Thanks: 345  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. 

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