Can four integer numbers $x$, $y$, $x-y$, $x+y$ be powerful numbers where $\gcd(x,y)=1$?

Question

Can four integer numbers $x$, $y$, $x-y$, $x+y$ be powerful numbers where $\gcd(x,y)=1$ ?

Answer 1

Using the infinitely many rational points on the curve $x^4-y^4=(2^4-1) y^2 z^2$, the smallest solution we found is with $x$ having 146 decimal digits (smaller solutions may exist).

Using the same curve, there are infinitely many solutions of very large size.

The solution in sage code:

def fourpowerful1():
    """
    
    x^2,y^2,x^2+y^2,x^2-y^2 all powerful
    Coming from the infinitely many points on the curve 
    x^4-y^4=(2^4-1) y^2 z^2
    Author:  Georgi Guninski, Thu Jul  4 10:21:11 AM UTC 2024
    """
    x_1=51203929262112985953504884979884461019907289047719199306830900384193163023961621160885389604169689049920413633464220309380754803019095566545430404
    y_1=5168374460623477253206993537636619974630340154166825903559471772312195920181455373026653181235106737210693085904835681664925203352727316123710721
    tp=x_1+y_1
    tm=x_1-y_1
    print("gcd=",gcd(x_1,y_1),x_1.is_square(),y_1.is_square(),(tp//5^3).is_square(),(tm//3^3).is_square(),"digits(x)=",len(str(x_1)) )
    print("x=",factor(x_1))
    print("y=",factor(y_1))
    print("x+y=",factor(tp))
    print("x-y=",factor(tm))

---

I think there are infinitely many solutions via rational points on elliptic curves, but the smallest solution is large.

Let $X=x^2,Y=y^2$, then we want $(X^2-Y^2)*(X^2+Y^2)=X^4-Y^4 =\text{powerful}$

In this question Elkies gave solution to similar problem $X^4+Y^4=\text{powerful}$: $427511122^4 + 1322049209^4=\text{powerful}$

Let $a_1,a_2,a_3$ be integers.

The projective curve $X^4-Y^4=a_1^3 (a_1 X+a_2 Y)^2 Z^2$ is genus one elliptic curve and it may have infinitely many points $(X,Y,Z)$ and I believe it may have infinitely many solutions with $(X,Y)$ coprime.

Someone may try to find points on the elliptic curve for some $a_i$.

Comment appears to suggest that abc implies finitely many solution, but I believe abc doesn't imply this.

As pointed by Peter Taylor, solutions to $X^4-Y^4=\text{powerful}$ exist like $10113607^4-4319999^4$, but in this particular case $X^2+Y^2$ is not powerful. It is a near miss and the only exponent of one is at the oddest prime $2$.

Answer 2

A smaller example than joro's, also obtained by the same elliptic-curve method, is $$ x = 7776485^2, \quad y = 6^3 439967^2, \quad x+y = 10113607^2, \quad x-y = 4319999^2. $$ I found it by starting from the near-miss $(x,y,x+y,x-y) = (25,24,49,1)$ and tripling the corresponding point $(X,Y,Z) = (5,1,7)$ on the curve $E_6: (X^2-6Y^2)(X^2+6Y^2)=Z^2$ to make $Y$ a multiple of 3. The appearance of $10113607^2$ and $4319999^2$ means that this solution could also have been obtained from the "near-miss" involving $10113607^4 - 4319999^4$ in Peter Taylor's comment.

Again we get infinitely many further solutions using the group law on $E_6$.

P.S. To make it easier to check the above solution:

x = 7776485^2
y = 6^3 * 439967^2
x+y == 10113607^2
x-y == 4319999^2

P.P.S. (added later) Here's what "tripling the point" looks like. I wrote $E_6$ as a curve in projective space with coordinates $X,Y,Z$ of weights $1,1,2$; setting $Y=1$ gives the affine curve $C: x^4 - 36 = z^2$ with a rational point $P_1: (x,z) = (5/2,7/4)$. We want a quadratic polynomial $Q(x)$ such that the graph of $z=Q(x)$ intersects $C$ at $P_1$ with multiplicity $3$; the fourth intersection will be the desired "tripled" point $P_3$ (actually $P_{-3}$, but that point serves the same purpose). We get $Q$ by expanding $z$ in a Taylor series about $x(P_1)$ to order 2: $$ z = \frac74 + \frac{125}{7} \Bigl(x - \frac52\Bigr) - \frac{27575}{7^3} \Bigl(x - \frac52\Bigr)^2 + \cdots $$ We thus set $Q(x) = -7^{-3} (27575 x^2 - 144000 x + 187056)$. Then $x^4 - 36 = Q(x)^2$ has the expected triple root at $x = 5/2$, and we calculate that the remaining root is $7776485 / 2639802.$ That's the $x$-coordinate of a new pair of rational points on $C$. The desired $X,Y$ are then the numerator and denominator of $x^2/6$.