stuck..

oh boy this is gonna be a long one

let f(x) = x² + ax + p (since leading coeff = 1 and f(0)=p)

given:

f(1) = 1 + a + p = 1/3

⇒ a + p = -2/3 … (1)

now key idea:

common real root of f(x)=0 and f(f(f(f(x))))=0

if r is a root of f(x)=0 ⇒ f(r)=0

then automatically:

f(f(r)) = f(0),

f(f(f(r))) = f(f(0)),

f(f(f(f(r)))) = f(f(f(0)))

so for r to be a common root, we need:

f(f(f(f(r)))) = 0 ⇒ f(f(f(0))) = 0

let’s simplify step by step

f(0) = p

f(f(0)) = f(p)

f(f(f(0))) = f(f(p))

so condition becomes:

f(f(p)) = 0

this means f(p) must be a root of f(x)=0

but roots of f are say α, β

so either f(p) = α or β

now smarter approach: since f is quadratic, try making p itself a root

assume p is a root ⇒ f(p)=0

⇒ p² + ap + p = 0

⇒ p(p + a + 1) = 0

p ≠ 0 ⇒ p + a + 1 = 0

⇒ a + p = -1 … (2)

but from (1): a + p = -2/3

but from (1): a + p = -2/3

contradiction ⇒ so p is NOT root

next, try: f(p) = 0’s other root behavior → use Vieta

roots α, β satisfy:

α + β = -a, αβ = p

we want f(p) to be one of α or β

compute f(p):

= p² + ap + p

= p(p + a + 1)

so f(p) = p(p + a + 1)

this must equal α or β

but αβ = p ⇒ suggests try matching f(p)=p

⇒ p(p + a + 1) = p

⇒ p + a + 1 = 1

⇒ a + p = 0 … (3)

nope

so try: f(p) = 0’s root ⇒ means f(p)=0

we already did → leads to contradiction

so the only consistent way is:

f(p) equals the OTHER root and structure forces:

p + a + 1 = -1/3

(using (1): a + p = -2/3)

so p + a + 1 = (-2/3 + 1) = 1/3

⇒ f(p) = p × (1/3) = [p/3](/forums/p/3)

this must be a root ⇒ so it satisfies equation:

([p/3](/forums/p/3))² + a([p/3](/forums/p/3)) + p = 0

multiply by 9:

p² + 3ap + 9p = 0

⇒ p² + p(3a + 9) = 0

⇒ p + 3a + 9 = 0 (since p≠0) … (4)

now solve (1) and (4):

(1): a + p = -2/3

(4): p + 3a + 9 = 0

subtract:

(4) - (1):

2a + 9 = 2/3

⇒ 2a = -25/3

⇒ a = -25/6

then p = -2/3 - a = -2/3 + 25/6 = 21/6 = 7/2

now f(-3):

= (-3)² + a(-3) + p

= 9 -3a + p

= 9 -3(-25/6) + 7/2

= 9 + 75/6 + 7/2

= 9 + 12.5 + 3.5

= 25