Uniquely square

We know that the highest degree of the polynomial is 2. So: $f(y)=y^2+cy+d$ Expanding $f(y)^2$ we get: $f(y)^2=y^4+2cy^3+(2d+c^2)y^2+2dcy+d^2$ Solving for $c$ and $d$ with what we know, we get that $c=1$ and $d=1$ . So we know that by plugging values in that $a=3$ and $b=1$ . Therefore $2a+9b=15$

Since $y^4 + 2y^3 + ay^2 + 2y +b$ can be written as $(f(y))^2$ , this means that it is a perfect square, so we can clearly state that $f(y)$ must be of a second degree (quadratic) form.

Therefore, let $f(y) = py^2 + qy + r$ , then

$(f(y))^2 = (py^2+qy+r)^2 = p^2y^4 + 2pqy^3 + y^2(q^2 + 2pr) + 2qry + r^2$ But $(f(y))^2 = y^4 + 2y^3 + ay^2 + 2y +b$

Comparing the coefficients in both the equations, we get -

$p^2 = 1$

$2pq=2 \Rightarrow q = \frac{1}{p}$

$2qr=2 \Rightarrow r = \frac{1}{q} = p$

Then $a = q^2 + 2pr = \frac{1}{p^2} + 2p^2 = 1 + 2 = 3$

And $b = r^2 = p^2 = 1$

Therefore, $2a + 9b = (2 \times 3) + (9 \times 1) = 15$

$f(y)^2 = (y^2+cy+d)^2$

$=y^4+2cy^3+(c^2+2d)y^2+2cdy+d^2$

Equating Coefficients:

$2c=2 \Rightarrow c=1$

$2cd=2 \Rightarrow 2 \times 1 \times d = 2 \Rightarrow d=1$

$a=(c^2+2d)=1^2 +2 \times 1 = 3$

$b=d^2=1^2=1$

$2a+9b= 2 \times 3 + 9 \times 1 =$ $\fbox{15}$

If y^4+2y^3+ay^2+2y+b= f(y)^2, then there is a polynomial f(y)= (Y^2+ny+sqrt(b)) where n is any real number. Squaring f(y) we get, (Y^2+nY+sqrt(b))(Y^2+nY+sqrt(b)) = (Y^4 + 2nY^3 + (n^2+ 2sqrt(b))Y^2 + 2nYsqrt(b) + b). Matching corresponding terms, we get n=1 and sqrt(b)=1. Therefore, a=3, and b=1. 2(3) + 9(1) = 15.

f(x)^2 must be /(y^2 + my + n)(y^2 + my +n)/

Multiplying this out we get /(y^4 + 2my^3 + (2n + m^2)y^2 + 2mny +n^2/).

This is equal to /(y^4 +2y^3 +ay^2 +2y +b/) for all values of y. From this we have m = 1, n = 1, and a = 3, b = 1.

/((2(3) + 9(1) = 15/)

it is the square of (y2+y+1)=y4+2y3+3y2+2y+1 ....2(3)+9(1)=15

(y^2+ry+s)^2 ==>

y^4 + 2r.y^3 + (r^2+2s).y^2 + 2rs.y + s^2=0

y^4 + 2y3 + ay2 + 2y + b = 0

Hence,

2r=2

r=1

2rs=2

s=1

a=r^2+2s=1^2+2*1=3

b=s^2=1^2=1

2a+9b=2(3)+9(1)=6+9=15