Polynomial Challenge: Show $f(5y^2)=P(y)Q(y)$

[anemone]

Given that $f(x)=x^4+x^3+x^2+x+1$. Show that there exist polynomials $P(y)$ and $Q(y)$ of positive degrees, with integer coefficients, such that $f(5y^2)=P(y)\cdot Q(y)$ for all $y$.

 

[MarkFL]

My solution:

Spoiler

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:

\(\displaystyle f\left(5y^2\right)=\left(25y^4+25y^3+15y^2+5y+1\right)\left(25y^4-25y^3+15y^2-5y+1\right)\)

 

[anemone]

My solution:

Spoiler

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:

\(\displaystyle f\left(5y^2\right)=\left(25y^4+25y^3+15y^2+5y+1\right)\left(25y^4-25y^3+15y^2-5y+1\right)\)

Good job, MarkFL! And thanks for participating!:)