# Proof of rational number

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.

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}

• 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.

##  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.)

##  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.

##  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.

##  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.

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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