Has equation $a^n+b^n=c^n+1$ infinite solutions with $n>2$?

Question

With $n=2$, it seems the solutions are infinitely many. For $n=3$ it seems number of solutions are limited; I could only find $(a, b, c)=(9, 10, 12)$. My question is; does this equation has infinite solutions with restriction:

$(a, ,c, b, n)=1$

For $n=2$, I found $(a, b, c)=(19, 89, 91)$ by brute force. There are probably more solutions.

For $n=3$, I found $(a, b, c)=(9, 10, 12)$ which does not fit the restriction. For $n>3$ I could not find any,May be my computer or my program is not suitable. Any help or counter example?

Update: The equation is consistent for $\bmod n$ if n is prime and (a.n)=, (b,n)=1 and (c, n)=1:

$a^n+b^n\equiv 2\bmod n\equiv c^n+1$

Identity $(1-9t^3)^3+(9t^4)^3+(3t-9t^4)^3=1$ shows this for $n=3$. There is no reason for the lack of solution for prime n.

Thanks to Tomita for giving following identity:

$$(1-9t^3)^3+(9t^4)^3+(3t-9t^4)^3=1$$

For $t=-1$, this identity gives $9^3+10^3=12^3+1$.

Also to Piquito for Identity:

$$(rX+sY)^2+(rY-sX)^2=(rX-sY)^2+(rY+sX)^2$$

@sirous FYI, using an Approach0 search, there's the closely related Solutions to $A^N+B^N=C^N \pm 1$. — John Omielan, Jul 31 '23 at 19:33
@W. Tanner, Yes , thy must be coprime too, like the example I gave for $n=2$ — sirous, Jul 31 '23 at 20:10
Without the restriction there are infinitely many with $(a,b,c)=(1,t,t)$ however they could be consider as "trivial". — Ataulfo, Jul 31 '23 at 20:19
We need don't forget the famous Mordell's conjecture, now Falting's theorem which says that all curve of genus greater than $1$ over $\mathbb Q$ has only a finite number of rational points. — Ataulfo, Jul 31 '23 at 21:30
Do you want $a, c, b, n$ to be pairwise coprime? If so, the notation $(a, b, c, n) = 1$ is incorrect, as it means that $\gcd(a, b, c, n) = 1$. — Denis Shatrov, Jul 31 '23 at 21:38
@sirous, Famous identity: $(1-9t^3)^3 + (9t^4)^3 + (3t-9t^4)^3 = 1$ — Tomita, Aug 02 '23 at 05:41
@Tomita, Thanks for excellent identity. I include it in my question. This identity is competent with what I found by computer for n=3 and t=-1, this indicates that $n \in $\Mathbb Z$, also for x=3 there is infinite solutions. — sirous, Aug 02 '23 at 11:19

Tito Piezas III · Answer 1 · 2023-08-21T14:42:46.607

I. The equation $a^3+b^3+c^3 = 1$ not only has infinitely many numerical solutions, it also has infinitely many polynomial solutions. The first one has already been given,

$$\small{(1 - 9 t^3)^3 + (9 t^4)^3 + (3t - 9 t^4)^3 = 1}$$

and the second is,

$$\small{(1 - 9t^3 + 648t^6 + 3888t^9)^3 + (-135t^4 + 3888t^{10})^3 + (3t - 81t^4 - 1296t^7 - 3888t^{10})^3 = 1}$$

which involve polynomials of degrees $4,10,16,$ and so on. The infinite process using Pell equations is described in this post.

II. Alternatively, one can use Pell equations directly. Given,

$$\small{(9p^2 - 176p q + 851q^2)^3 + (-12 p^2 + 224 p q - 1052 q^2)^3 + (10 p^2 -180 p q + 818 q^2)^3 = (p^2 - 85 q^2)^3}$$

$$\small{(9 p^2 - 312 p q + 2727 q^2)^3 + (-8 p^2 + 270 p q - 2238 q^2)^3 + (-6 p^2 + 222 p q - 2088 q^2)^3 = (p^2 - 321 q^2)^3}$$

One just solves the Pell equation,

$$p^2 - 85 q^2 = 1$$ $$p^2 - 321 q^2 = 1$$

and you are done.

P.S. And since there are infinitely many of these quadratic parameterizations, then there are also infinitely many Pell equations one can choose from.

Ataulfo · Answer 2 · 2023-07-31T22:34:30.777

►It seems to me that in fact for $n=2$ there are infinitely many solutions but they all are "trivial" (in the sens of as see at comment above). In fact the general solution of the equation $a^2+b^2=c^2+d^2$ is given by the identity with four parameters $$(rX+sY)^2+(rY-sX)^2=(rX-sY)^2+(rY+sX)^2\hspace 2cm(*)$$ so for each choice of $r,s$ coprimes we have the restriction $$rY+sX=1\hspace 2cm(**)$$ which have an infinite number of integer solutions. However the standard form of these last solutions leads for each choice of $r,s$ and each solution of $(**)$ to a solution of $(*)$ which is of the form $(a,b,c)=(1,t,t)$ as we can verify.

►►The following, I believe, is a plausible first step to a proof of the impossibility for $n$ even i leave to the O.P. to finish if he want to.

If $n$ is even then the curve $x^n+y^n=k$ where $k=c^n+1$ is closed and $(\pm x)^n+(\pm y)^n=k$ so it is enough to study the curve in the first quadrant, By convexity, the arc of the curve defined by the "trivial" solutions $(c,1)$ and $(1,c)$ is inside the triangle of vertex $(c,1),(1,c),(c,c)$ and we can deduce that this arc does not have any other integer point that the two trivial solutions.

Thanks for your answer . I included your Identity in my question. — sirous, Aug 02 '23 at 11:13

Has equation $a^n+b^n=c^n+1$ infinite solutions with $n>2$?

2 Answers2

Linked