MarkFL said:My solution:
We find that:
\(\displaystyle f\left(5y^2\right)=625y^8+125y^6+25y^4+5y^2+1\)
Let us assume then that this can be factored into two quartics as follows:
\(\displaystyle 625y^8+125y^6+25y^4+5y^2+1=\left(25y^4+ay^3+by^2+cy+1\right)\left(25y^4-ay^3+by^2-cy+1\right)\)
Let us further assume that all of $a,b,c$ are positive.
Expansion of the right side yields:
\(\displaystyle 625y^8+125y^6+25y^4+5y^2+1=625y^8+\left(50b-a^2\right)y^6+\left(b^2-2ac+50\right)y^4+\left(2b-c^2\right)y^2+1\)
Equating coefficients gives the non-linear system:
\(\displaystyle 50b-a^2=125\)
\(\displaystyle 2ac-b^2=25\)
\(\displaystyle 2b-c^2=5\)
The first and third give:
\(\displaystyle 2b=5+\left(\frac{a}{5}\right)^2=5+c^2\implies a=5c\)
Substituting into the second equation, we obtain:
\(\displaystyle 10c^2-b^2=25\)
Multiplying the 3rd equation by 10, we find:
\(\displaystyle 20b-10c^2=50\)
Adding the last two results, we eliminate $c$ to obtain:
\(\displaystyle -b^2+20b=75\)
\(\displaystyle b^2-20b+75=0\)
\(\displaystyle (b-5)(b-15)=0\)
We then find that only the root $b=15$ allows $a$ and $c$ to be integers:
\(\displaystyle a=25,\,c=5\)
Hence:
\(\displaystyle f\left(5y^2\right)=\left(25y^4+25y^3+15y^2+5y+1\right)\left(25y^4-25y^3+15y^2-5y+1\right)\)
A polynomial is an algebraic expression that consists of mathematical operations such as addition, subtraction, and multiplication, using variables and constants. It can have multiple terms with different powers of the variables, but it does not include operations like division, radicals, or fractions.
The Polynomial Challenge is a mathematical problem that involves showing that an equation is true for all values of a variable. In this case, the challenge is to prove that the equation $f(5y^2)=P(y)Q(y)$ holds for all possible values of the variable y.
The notation $f(5y^2)$ means that the function f is being evaluated for a specific value of the variable, which in this case is 5y^2. This means that the value of the function will depend on the value of 5y^2, and not just y.
In this equation, $P(y)$ and $Q(y)$ represent polynomial expressions. They could be any type of polynomial, such as a linear, quadratic, or cubic polynomial. The challenge is to show that the equation holds for any type of polynomial expression.
To prove the Polynomial Challenge, one must use mathematical techniques such as substitution, factoring, and algebraic manipulation. The key is to show that the equation holds true for all values of y, which can be achieved by manipulating the equation and showing that it simplifies to a true statement for all values of y.