Can anyone post a hint please?

Well, if such a curve exists, then translate the curve down s.t. the center of the square is at (0,0), to get another monic cubic curve. Such a transformation preserves length, so after we find the length of the side of such a square, we can just translate back to the original curve. If the point (x,y) is a point, then so is (-y,x) (-x,-y) (y,-x). $ x^3 + ax^2 + bx + c = y$ $ ( - x)^3 + a( - x)^2 + b( - x) + c = - y$ Summing the 2 up, we have $ 2ax^2 + 2c = 0$ $ ( - y)^3 + a( - y)^2 + b( - y) + c = x$ $ y^3 + ay^2 + by + c = - x$ Summing up the 2, we have $ 2ay^2 + 2c = 0$ Now, if a is not equal to 0, we have x,y are negatives of each other, or are equal to each other, since they are both solutions to $ 2at^2 + 2c = 0$ Therefore, $ (x,x)$ and $ (x, - x)$ would both have to be points, but since cubics are one-to-one, x=-x, so $ x = 0$, so no square exists. Hence, a=0, and therefore c=0. The set of equations reduces to: $ x^3 + bx = y$ $ y^3 + by = - x$ x=0 or y=0 give degenerate squares, so there has to exist a non-zero pair (x,y) which satisfy the system. If b is positive then if x is positive, y is positive, so -x is positive, contradiction. If b is positive then if x is negative, y is negative, so -x is negative, contradiction. Hence, b is either 0 or is a negative number. Anyone can finish?

Rofler wrote: Well, if such a curve exists, then translate the curve down s.t. the center of the square is at (0,0), to get another monic cubic curve. Such a transformation preserves length, so after we find the length of the side of such a square, we can just translate back to the original curve. If the point (x,y) is a point, then so is (-y,x) (-x,-y) (y,-x). $ x^3 + ax^2 + bx + c = y$ $ ( - x)^3 + a( - x)^2 + b( - x) + c = - y$ Summing the 2 up, we have $ 2ax^2 + 2c = 0$ $ ( - y)^3 + a( - y)^2 + b( - y) + c = x$ $ y^3 + ay^2 + by + c = - x$ Summing up the 2, we have $ 2ay^2 + 2c = 0$ Now, if a is not equal to 0, we have x,y are negatives of each other, or are equal to each other, since they are both solutions to $ 2at^2 + 2c = 0$ Therefore, $ (x,x)$ and $ (x, - x)$ would both have to be points, but since cubics are one-to-one, x=-x, so $ x = 0$, so no square exists. Hence, a=0, and therefore c=0. The set of equations reduces to: $ x^3 + bx = y$ $ y^3 + by = - x$ x=0 or y=0 give degenerate squares, so there has to exist a non-zero pair (x,y) which satisfy the system. If b is positive then if x is positive, y is positive, so -x is positive, contradiction. If b is positive then if x is negative, y is negative, so -x is negative, contradiction. Hence, b is either 0 or is a negative number. Anyone can finish? continue with Rofler's solution, now we have non-zero $x$, $y$ $ x^3 + bx = y$ $ y^3 + by = - x$ but we need to use the condition that there is "exactly one" square on the cubic curve, which is equivalent to there is exactly one solution (x, y) in the 1st quadrant satisfy the above equations. Plot the cubic curves we can figure out this implies the two curves tangent at (x, y). then $ (3x^2 + b)dx = dy$ $ (3y^2 + b)dy = - dx$ cross out we have $ (3x^2 + b)(3y^2 + b) + 1 = 0 $ now there are three equations and three variables $x,y,b$ so we can, in principle, solve the equations

the details are as follows rewrite $ (3x^2+b)(3y^2+b)+1 = 0 $ as $ x^2 y^2 = -(\frac{b(x^2+y^2)}{3} + \frac{b^2+1}{9}) $ ...(P) square $ x^3+bx = y $ and $ y^3+by =-x $ : $ x^6 + 2bx^4 + b^2 x^2 = y^2 $ ...(M) $ y^6 + 2by^4 + b^2 y^2 = x^2 $ ...(N) (M) + (N) : $ x^2+y^2 = (x^6+y^6) + 2b(x^4+y^4) + b^2(x^2+y^2) $ $ = (x^2+y^2)^3 - 3x^2 y^2(x^2+y^2) + 2b(x^2+y^2)^2 - 4bx^2 y^2 + b^2(x^2+y^2) $ $ = (x^2+y^2)^3 + (b(x^2+y^2) + \frac{b^2+1}{3})(x^2+y^2) + 2b(x^2+y^2)^2 $ $ + 4b(\frac{b(x^2+y^2)}{3} + \frac{b^2+1}{9}) + b^2(x^2+y^2) $ [ insert (P)] $ = (x^2+y^2)^3 + 3b(x^2+y^2)^2 + \frac{8b^2+1}{3}(x^2+y^2) + \frac{4b(b^2+1)}{9}$ rearrange and denote $R = x^2+y^2$ : $ R^3 + 3bR^2 + \frac{(8b^2-2)R}{3} + \frac{4b(b^2+1)}{9} = 0 $ ...(U) (M) - (N) : $ (x^6-y^6) + 2b(x^4-y^4) + (b^2+1)(x^2-y^2) = 0 $ divides $x^2-y^2$ $ 0 = (x^4+x^2 y^2+y^4) + 2b(x^2+y^2) + (b^2+1) $ $ = (x^2+y^2)^2 - x^2 y^2 + 2b(x^2+y^2) + (b^2+1) $ $ = (x^2+y^2)^2 + \frac{b(x^2+y^2)}{3} + \frac{b^2+1}{9} + 2b(x^2+y^2) + (b^2+1) $ [ insert (P)] $ = (x^2+y^2)^2 + \frac{7b(x^2+y^2)}{3} + \frac{10(b^2+1)}{9} $ rearrange $ R^2 + \frac{7}{3}bR + \frac{10(b^2+1)}{9} = 0 $ ...(V) (U) x $5-$ (V) x $2b$ : $ 5R^3 + 13bR^2 + \frac{26}{3}b^2 R - \frac{10}{3}R = 0 $ ...(W) (W) x $\frac{1}{R}+$ (V) x $3$ : $ 8R^2 + 20bR + 12b^2 = 0 $ $ (2R+3b)(R+b) = 0 $ if $R+b=0$, insert back to (V) get $R=\sqrt 5, b=-\sqrt 5$ $x^2+y^2=\sqrt 5, x^2 y^2=1,$ then $\{x,y\}=\{\sqrt{\frac{\sqrt 5 + 1}{2}},\sqrt{\frac{\sqrt 5 - 1}{2}}\}$ insert to $ x^3+bx = y $ will get into contradiction so $2R+3b=0$, insert back to (V) get $R=3\sqrt 2, b=-2\sqrt 2$ the square has sides of length $\sqrt{2(x^2+y^2)}=\sqrt{2R}=\sqrt[4]{72}$

Maybe a solution with a bit less algebra. Shift the coordinate system so that the center of the square is the origin and let $P(x)=x^3+ax^2+bx+c$. Then $P(x)+P(-x)=0$ for the four vertices of the square. Since this is a quadratic, it must be identically $0$. Hence, $a=c=0$. Thus, $y=x^3+bx$. Rotating this by $90^\circ$ around the origin leads to the curve $x=-y^3-by$. Since exactly one square is possible, the two cubics have to be tangent to another. Hence, by substituting $y=x^3+bx$ into the other equation (and dividing out the solution $x=0$), we must get a polynomial with four double roots. Hence, $0=x+y^3+by=x+(x^3+bx)^3+b(x^3+bx)=x(x^8+3bx^6+3b^2x^4+(b^3+b)x^2+b^2+1)$. Setting the second factor equal to $(x^4+a_3x^3+a_2x^2+a_1x+a_0)^2$ and comparing coefficients leads to $a_1=a_3=0$, $3b=2a_2, 3a_0=b^2+1,a_2^2+2a_0=3b^2$. This implies $b=\pm 2\sqrt 2$. Now $b<0$, otherwise $P$ would be increasing and the two curves would only intersect at $(0,0)$. Hence, $b=-2\sqrt 2$. Then the factorization becomes $(x^4-3\sqrt 2x^2+3)^2$. If the coordinates of the vertices are $(x_i,y_i),1\leq i\leq 4$ and the square has sidelength $a$, then by the quadratic formula $a^2=\frac 12(x_1^2+y_1^2+\hdots+x_4^4+y_4^2)=x_1^2+x_2^2+x_3^2+x_4^2=4\cdot\frac{3\sqrt 2}{2}=\sqrt {72}$, as desired.