i got a different answer that checks out, so i'm curious what i may have missed...

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Let P represent the product of roots and S the sum of roots.

For clarity, let the original quadratic and conditions be defined as:
y = αx^2 + 	βx + γ
  {  P = 1
  {  S = α
  { βγ = -4

By using the quadratic equation, the roots of this quadratic are:
[-b ± sqrt(b^2 - 4ac)] / [2a]

P = ( [-β + sqrt(β^2 - 4αγ)] / [2α] )( [-β - sqrt(β^2 - 4αγ)] / [2α] )
  = (-β + sqrt(β^2 - 4αγ))(-β - sqrt(β^2 - 4αγ)) / (2α)(2α)
  = (β^2 - sqrt(β^2 - 4αγ)^2) / 4(α)^2
  = (β^2 - (β^2 - 4αγ)) / 4(α)^2
  = 4αγ / 4(α)^2
P = γ/α.


S = ( [-β + sqrt(β^2 - 4αγ)] / [2α] ) + ( [-β - sqrt(β^2 - 4αγ)] / [2α] )
  = -2β / 2α
  = -β/α.

So, the system of equations can be written and labelled as:
(1)  {  γ/α = 1  <-> γ = α
(2)  { -β/α = α  <-> β = -(α)^2
(3)  {   βγ = -4 <-> β = -4/γ    [ {α,	β,γ} in Z+ ]

(2)=(3):      -(α)^2 = -4/γ
    (1):      -(α)^2 = -4/α
         -(α)^2 * -α = 4
               (α)^3 = 4
                   α = 4^(1/3).

α --> (2): β = -(α)^2
           β = -(4^(1/3))^2
           β = -(4)^(2/3).

α --> (1): γ = α
           γ = 4^(1/3).

For simplicity, let ρ = 4^(1/3). Then: α = ρ; β = -(ρ)^2; γ = ρ

So, the original quadratic is:
y = ρx^2 - 2(ρ^2)x + ρ.
`-----------------------`

Check:

Condition 1:
P = 1 ?= γ/α
    1 ?= ρ/ρ
    1 ✓= 1.

Condition 2:
S = α ?= -β/α
    ρ ?= -[-(ρ)^2] / ρ
    ρ ?= (ρ)^2 / ρ
  ρ^2 ✓= ρ^2.

Condition 3:
        βγ ?= -4
-(ρ)^2 * ρ ?= -4
 (ρ)^2 * ρ ?= 4
       ρ^3 ?= 4
 4^(1/3)^3 ?= 4
   4^(3/3) ?= 4
         4 ✓= 4.