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This is my attempt to give a more motivated version of darij grinberg's argument.

Let's try to work in the greatest possible degree of generality. The equation $X^3 + aX^2 + b X + c$ implies the equations in the roots $\alpha_1,\alpha_2,\alpha_3$,

$$\alpha_1 + \alpha_2 + \alpha_3 = -a,$$

$$\alpha_1\alpha_2+ \alpha_2\alpha_3 +\alpha_1\alpha_3=b,$$

$$\alpha_1\alpha_2\alpha_3 = -c.$$

If we imagine that $a$ and $b$ are fixed and $c$ is varying, we can drop the last equation. Because we have two equations in three variables, we can deduce one equation in any two of the variables, say $\alpha_1$ and $\alpha_2$.

When does that equation force their product to be a square?

Writing $\alpha_3 = -a-\alpha_1-\alpha_2$, the equation in $\alpha_1,\alpha_2$ is

$$\alpha_1\alpha_2 = (\alpha_1+\alpha_2) (a + \alpha_1 + \alpha_2) + b. $$

Thus, whenever $b=a^2/4$, the formula on right side is $(\alpha_1 + \alpha_2 + a/2)^2$ and $\alpha_1\alpha_2$ is forced to be a square if $F$ if $\alpha_1,\alpha_2 \in F$.

EDIT: Following a clever observation of user44191 in the comments:

If $f(x)$ is a monic polynomial, and $c$ a number, then the polynomial $xf(x)^2+c$ has a similar property to your example (the case $f(x)=x+1/2$). Indeed, we have $x = \frac{-c}{f(x)^2}$ so

• If $-c$ is a nonzero square then all rational roots are squares.
• if $-c$ is a nonsquare then all rational roots are not square, but their ratios are square.
• If $-c$ is zero then one root is zero and the rest are double (this doesn't really fit the pattern).

This produces polynomials of odd degree. For even degree examples, we can do $f(x)^2+cx$. This gives $x =\frac{ f(x)^2}{-c}$ so we have the same thing except if $-c$ is zero than all roots are double, and there is a special case if $f(0)=0$.

So we have many examples of polynomials of this type.

(See the edit history for an earlier argument, special to the case of degree 3 polynomials, if desired. This was inspired by darij grinberg's answer, and that earlier answer inspired user44191's comment, so both of them are partially responsible for this solution.)

This is my attempt to give a more motivated version of darij greenberg's argument.

Let's try to work in the greatest possible degree of generality. The equation $X^3 + aX^2 + b X + c$ implies the equations in the roots $\alpha_1,\alpha_2,\alpha_3$,

$$\alpha_1 + \alpha_2 + \alpha_3 = -a,$$

$$\alpha_1\alpha_2+ \alpha_2\alpha_3 +\alpha_1\alpha_3=b,$$

$$\alpha_1\alpha_2\alpha_3 = -c.$$

If we imagine that $a$ and $b$ are fixed and $c$ is varying, we can drop the last equation. Because we have two equations in three variables, we can deduce one equation in any two of the variables, say $\alpha_1$ and $\alpha_2$.

When does that equation force their product to be a square?

Writing $\alpha_3 = -a-\alpha_1-\alpha_2$, the equation in $\alpha_1,\alpha_2$ is

$$\alpha_1\alpha_2 = (\alpha_1+\alpha_2) (a + \alpha_1 + \alpha_2) + b. $$

Thus, whenever $b=a^2/4$, the formula on right side is $(\alpha_1 + \alpha_2 + a/2)^2$ and $\alpha_1\alpha_2$ is forced to be a square if $F$ if $\alpha_1,\alpha_2 \in F$.

This is my attempt to give a more motivated version of darij grinberg's argument.

Let's try to work in the greatest possible degree of generality. The equation $X^3 + aX^2 + b X + c$ implies the equations in the roots $\alpha_1,\alpha_2,\alpha_3$,

$$\alpha_1 + \alpha_2 + \alpha_3 = -a,$$

$$\alpha_1\alpha_2+ \alpha_2\alpha_3 +\alpha_1\alpha_3=b,$$

$$\alpha_1\alpha_2\alpha_3 = -c.$$

If we imagine that $a$ and $b$ are fixed and $c$ is varying, we can drop the last equation. Because we have two equations in three variables, we can deduce one equation in any two of the variables, say $\alpha_1$ and $\alpha_2$.

When does that equation force their product to be a square?

Writing $\alpha_3 = -a-\alpha_1-\alpha_2$, the equation in $\alpha_1,\alpha_2$ is

$$\alpha_1\alpha_2 = (\alpha_1+\alpha_2) (a + \alpha_1 + \alpha_2) + b. $$

Thus, whenever $b=a^2/4$, the formula on right side is $(\alpha_1 + \alpha_2 + a/2)^2$ and $\alpha_1\alpha_2$ is forced to be a square if $F$ if $\alpha_1,\alpha_2 \in F$.

This is my attempt to give a more motivated version of darij greenberg's argument.

Let's try to work in the greatest possible degree of generality. The equation $X^3 + aX^2 + b X + c$ implies the equations in the roots $\alpha_1,\alpha_2,\alpha_3$,

$$\alpha_1 + \alpha_2 + \alpha_3 = -a$$

$$\alpha_1\alpha_2+ \alpha_2\alpha_3 +\alpha_1\alpha_3=-b$$

$$\alpha_1\alpha_2\alpha_3 = -c$$

If we imagine that $a$ and $b$ are fixed and $c$ is varying, we can drop the last equation. Because we have two equations in three variables, we can deduce one equation in any two of the variables, say $\alpha_1$ and $\alpha_2$.

When does that equation force their product to be a square?

Writing $\alpha_3 = -a-\alpha_1-\alpha_2$, the equation in $\alpha_1,\alpha_2$ is

$$\alpha_1\alpha_2 = (\alpha_1+\alpha_2) (a + \alpha_1 + \alpha_2) - b $$

Thus, whenever $b=-a^2/4$, the formula on right side is $(\alpha_1 + \alpha_2 + a/2)^2$ and $\alpha_1\alpha_2$ is forced to be a square if $F$ if $\alpha_1,\alpha_2 \in F$.

This is my attempt to give a more motivated version of darij greenberg's argument.

Let's try to work in the greatest possible degree of generality. The equation $X^3 + aX^2 + b X + c$ implies the equations in the roots $\alpha_1,\alpha_2,\alpha_3$,

$$\alpha_1 + \alpha_2 + \alpha_3 = -a,$$

$$\alpha_1\alpha_2+ \alpha_2\alpha_3 +\alpha_1\alpha_3=b,$$

$$\alpha_1\alpha_2\alpha_3 = -c.$$

If we imagine that $a$ and $b$ are fixed and $c$ is varying, we can drop the last equation. Because we have two equations in three variables, we can deduce one equation in any two of the variables, say $\alpha_1$ and $\alpha_2$.

When does that equation force their product to be a square?

Writing $\alpha_3 = -a-\alpha_1-\alpha_2$, the equation in $\alpha_1,\alpha_2$ is

$$\alpha_1\alpha_2 = (\alpha_1+\alpha_2) (a + \alpha_1 + \alpha_2) + b. $$

Thus, whenever $b=a^2/4$, the formula on right side is $(\alpha_1 + \alpha_2 + a/2)^2$ and $\alpha_1\alpha_2$ is forced to be a square if $F$ if $\alpha_1,\alpha_2 \in F$.