Prove that the only solution to $X^4+4=pY^4$. is $X=\pm 1,Y=\pm 1,p=5$ for $p \ge 3$ prime.

Question

Let $p$, be a prime, $p\ge 3$. Consider the equation $X^4+4=pY^4$. Prove that the only solution is $X=\pm 1,Y=\pm 1,p=5$

Hint: $x^4+4=(x^2-2x+2)(x^2+2x+2)$. I've tried Fermat's Little Theorem and the linear Diophantine Equation, but it doesn't work. I have been taught Congruence

Answer 1

Suppose $x,y\geq 0$ and let $g:= \gcd(x^2-2x+2,x^2+2x+2)$

Let prime $q\mid g$ then $q\mid 4x$. If $q\ne 2$ then $q\mid x$ but then $$q\mid x^2+2x \implies q\mid 2 \implies q=2$$

So the only prime which can divide $ \gcd(x^2-2x+2,x^2+2x+2)$ is 2.

• $\color{red}{Case\; 1}$: $2\mid g$ then $x$ is even, so $x=2z$ then we see $4\nmid g$ and thus $g=2$. But then $2\mid y$ so $16\mid 4(4z^4+1)$ which is impossible.
• $\color{red}{Case\; 2}$: $2\nmid g$, so $g=1$, then $$x^2-2x+2 = pa^4 \;\;\;\wedge \;\;\;x^2+2x+2=b^4$$ where $a,b$ are relatively prime or $$x^2-2x+2 = a^4 \;\;\;\wedge \;\;\;x^2+2x+2=pb^4$$

• $\color{red}{Case\; 2.1}$: $$(x+1)^2<x^2+2x+2 < (x+2)^2$$ so no solution.

  • $\color{red}{Case\; 2.2}$: If $x>1$ then $$(x-1)^2<x^2-2x+2 < x^2$$ so no solution. So $x=1$ ...

Answer 2

Notice that $\text{gcd}(X^2+2X+2,X^2-2X+2)=\text{gcd}(X^2+2X+2,4X)=2$ or $1$. If $2|X$ then 2 must also divide $Y$ and thus: $4X^4+1=4Y^4$ a contradiction.

Then $\text{gcd}(X^2+2X+2,X^2-2X+2)=1$ and since $(X^2+2X+2)(X^2-2X+2)=pY^4$ this implies:

1) $X^2+2X+2={Y_1}^4$ and $X^2-2X+2=p{Y_2}^4$ or

2) $X^2+2X+2=p{Y_1}^4$ and $X^2-2X+2={Y_2}^4$.

Let us consider the first case: write ${Y_1}^4-(X+1)^2=1$ and observe that $({Y_1}^2-X-1)({Y_1}^2+X+1)=1 \implies {Y_1}^2-X-1={Y_1}^2+X+1=1$. Can you take it from here for both cases?

Answer 3

The hint essentially solves the problem.

First remark that $X,Y$ must both be odd (if one is even then so must the other be, but $16\,\nmid\,4$). It then follows quickly that the two factors on the left are relatively prime.

Since we have $$(X^2+2X+2)(X^2-2X+2)=pY^4$$ it must be the case that one or the other of the two left hand factors is a fourth power. Let's say $X^2+2X+2$ is a fourth power. Noting that $X^2+2X+1$ is a square we'd then have two consecutive squares, but the only two such are $0,1$. Similarly with $X^2-2X+2$. Simple case work then finishes the problem.