MHB Unraveling the Mystery: Solving for Real Values in a Complex Equation

[anemone]

Hi MHB,

Initially, I thought this is another boring high school mathematics problem, but when I started to work on it, I realized I was beaten by it, with equations in variables $a$, $b$, $p$ and $q$ to which I don't see a clear way to find the values for them.

Problem:

Find all real $p$, $q$, $a$ and $b$ such that we have $$(2x-1)^{20}-(ax+b)^{20}=(x^2+px+q)^{10}$$ for all $x$.

Attempt:

After expanding both sides of the equation using wolfram, I get

$$(1048576-a^{20})x^{20}-(10485760+20a^{19}b)x^{19}+(49807360-190a^{18}b^2)x^{18}+\cdots-(1140a^3b^{17}+9120)x^{3}$$
$$
+(-190a^2b^{18}+760)x^{2}-(20ab^{19}+40)x+1-b^{20}=x^{20}+10px^{19}+(10q^9+45p^2q^8)x^{2}+10pq^9x+q^{10}$$

And I ended up getting extremely messy equations where solving for the values for $a$, $b$, $p$ and $q$ seems impossible by equating the coefficient of $x^{20}$, $x^{19}$, $x^{18}$, $x^{3}$, $x^{2}$, $x$ and the constant...

Could anyone help me with this problem?

Thanks in advance.

 

[alyafey22]

Some solutions would be obtained by letting

$$a=2,b=-1$$

and finding the roots for

$$x^2+px+q=0$$

 

[kaliprasad]

Because it is true for all x you can choose any value of x

setting x = 1/2 shall make 1st term = 0

we get

- (b+1/2a)^20 = (1/4 + 1/2p+ q) ^ 10

LHS <=0 and RHS >= 0 so both are 0

So b =- 1/2a and 1/2p + q + 1/4 = 0

So a = - 2b and 2p + 2q + 1 = 0

by putting suitable values of x you can proceed.

 

[kaliprasad]

Because it is true for all x you can choose any value of x

setting x = 1/2 shall make 1st term = 0

we get

- (b+1/2a)^20 = (1/4 + 1/2p+ q) ^ 10

LHS <=0 and RHS >= 0 so both are 0

So b =- 1/2a and 1/2p + q + 1/4 = 0

So a = - 2b and 2p + 2q + 1 = 0

by putting suitable values of x you can proceed.

put x = 0 to get 1= b^20 + q^ 10

so b= (sin t)^(1/10) , q = (cos t)^(1/5)

a = -2 (sin t)^(1/10), p = - ( 1+ (cos t)^(1/5))/2

should be the solution for some t

unless I have missed out something

 

[Opalg]

Find all real $p$, $q$, $a$ and $b$ such that we have $$(2x-1)^{20}-(ax+b)^{20}=(x^2+px+q)^{10}$$ for all $x$.

Compare coefficients of $x^{20}$ to get $2^{20} - a^{20} = 1$. So $a = \pm (2^{20} - 1)^{1/20}$. As kaliprasad points out, $b = -\frac12a$, so $b = \mp\frac12(2^{20} - 1)^{1/20}$. Therefore $ax+b = \pm(2^{20} - 1)^{1/20}\bigl(x-\frac12\bigr)$, and $(ax+b)^{20} = (2^{20} - 1)\bigl(x-\frac12\bigr)^{20}$. Thus $$(2x-1)^{20}-(ax+b)^{20}= 2^{20}\bigl(x-\tfrac12\bigr)^{20} - (2^{20} - 1)\bigl(x-\tfrac12\bigr)^{20} = \bigl(x-\tfrac12\bigr)^{20} = \bigl(x^2- x + \tfrac14\bigr)^{10}.$$ So we must take $p=-1$ and $q=\frac14.$

So the solution is $a = \pm (2^{20} - 1)^{1/20}$, $b = \mp\frac12(2^{20} - 1)^{1/20}$, $p=-1$, $q=\frac14.$

 

[anemone]

Thank you all for the replies...I greatly appreciate all the helps!:)