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Prove that if $a$ and $b$ are rational numbers satisfying $a^5+b^5=2a^2b^2$, then $1-ab$ is the square of a rational number.

I am just a Year 2 student learning Abstract Algebra. This problem is a challenging one that my teacher gives us. However, I have no ideas about how to solve it. I tried to represent rational numbers by $p/q$ ($p$ and $q$ are both integers) but failed. I would appreciate it very much if anyone can help me.

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I don't know abstract algebra. I'm just going to use algebra precalculus.

  • If $b=0$, then the statement is correct.

Let $\dfrac {a}{b}=x, b≠0$, then we have

$$\begin{align}&a^5+b^5=2a^2b^2\\ \implies &a\times x^4+b=2x^2 \\ \implies &a \left(x^2\right)^2-2x^2+b=0\\ \implies &\Delta=1-ab=T^2, T\in\mathbb Q.\end{align}$$


I add additional information to the answer:

  • If $a,b$ is rational and $b≠0$, then $\dfrac {a}{b}$ is also rational.

  • Our equation is a quadratic equation with respect to $x^2=\left(\dfrac {a}{b}\right)^2$.

  • In order for the root of a polynomial equation whose coefficients are rational to be rational, the polynomial discriminant must also be a perfect square of the rational number.

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[1] Roots of quadratic equations

We'll need to recall a few facts about the roots of quadratic equations. For any quadratic equation $ax^2 + bx + c = 0$, the roots are $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

When will the roots of this equation be rational? Suppose $a$ and $b$ are rational. If the square root $\sqrt{b^2-4ac}$ of the discriminant is also rational, then note that the roots will be rational. The converse is also true: if the roots are rational, then $\sqrt{b^2-4ac}$ must be rational.

(You can prove these if you know that sums and products and ratios of rational numbers are rational.)

[2] The given equation

You can make the given equation $a^5 + b^5 = 2a^2b^2$ look like a quadratic equation. First rewrite it as $a(a^4) - 2a^2b^2 + b(b^4) = 0$ to expose even exponents. (This is the main trick to get the solution!) Divide by $b^4$ (this is always possible as long as $b\neq 0$) to get $a(a/b)^4 - 2(a/b)^2 + b = 0$.

Now this expression looks like the quadratic equation: $ax^2 - 2x + b = 0$, evaluated at $x=(a/b)^2$. So our given equation "Assume that $a^5 + b^5 = 2a^2b^2$" is equivalent to:

Assume that the quadratic equation $ax^2 - 2x + b = 0$ has $x=(a/b)^2$ as a root.

[3] The discriminant

We have determined that, by assumption, the equation $ax^2 - 2x + b = 0$ has $x=(a/b)^2$ as a root. This root is a rational number. Therefore (by #1), the square root of the discriminant is rational. That is, $\sqrt{(-2)^2 - 4ab}$ is rational. Dividing by 2, we have that $\sqrt{1-ab}$ is rational— which was to be shown.

[4] The edge case $b=0$

In the above manipulation, we divided by $b$. This works in every case except when $b=0$. Let's consider that case now: When $b=0$, then we are asked to prove that $1-ab = 1-0 = 1$ is the square of a rational number, which is straightforwardly true.

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Here is an amusingly overpowered proof. Note that the plane curve $C \subset \mathbb{A}^2$ cut out by $a^5+b^5=2a^2b^2$ has genus $0$. In particular it must admit a rational parametrisation $\phi : \mathbb{P}^1 \to C$, that is, we can write $a$ and $b$ as rational functions of one variable $t$ and then simply check that $1 - ab$ is a square in $\mathbb{Q}(t)$. Because $\phi$ is rational, there may be a finite number of points in the base locus of $\phi^{-1}$ to deal with, however as we'll see these are not an issue.

For the lazy magma has nice routines to parametrise rational curves.

A2<a,b> := AffineSpace(Rationals(), 2);

f := a^5+b^5-2*a^2*b^2;
C := Curve(A2,f);

C2 := ProjectiveClosure(C);
p := C2![1,1,1];

phi := Parametrization(C2, p);
aa, bb, cc := Explode(DefiningPolynomials(phi));
aa := aa/cc;
bb := bb/cc;

IsSquare(1- aa*bb);

BasePoints(Inverse(phi)); 
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