Is there a solution to AN+BN=CN±1 where A,B,C,N∈N, such that A,B,C>1,N≥4 and gcd(A,B)=gcd(B,C)=gcd(A,C)=1?

This question was inspired by this one: AX+BY=CZ±1 Beal’s conjecture “almost” solutions

A related question was asked here before.

Noam Elkies has published a paper on this subject. The table on page 15 of the paper and his extended table here have some interesting results. If I understood his work correctly, there are no solutions for 4≤N≤20 and C<223=8,388,608.

I also found a MathWorld article on 4th order diophantine equations, but it doesn't cover equations of this form. I also checked that the OEIS has no sequences for such Fermat near misses.

**Answer**

This exact question was a summer research project of mine when I was a second-year undergraduate in 2013. It was inspired by infinitely many solutions parametrized by Ramanujan in the n=3 case. I didn't make much progress, but I can show you a conditional result that I found, and some slight progress in the even case (with help from Noam Elkies, if I recall correctly). I haven't look at this in 7 years, so please verify independently if the propositions are true.

**Abstract.** Srinivasa Ramanujan wrote down in his Lost Notebook infinitely many non-trivial examples of integers x,y,z such that x^3+y^3=z^3\pm 1, for each sign. Aside from his miraculous construction, the result itself is remarkable as it shows that the smallest perturbation of Fermat's famed equation of degree 3, which has been known since the time of Euler to have no solutions, has infinitely many solutions. We study the natural generalization a^n+b^n=c^n\pm 1, which we call the Fermat-Ramanujan equation.

**Definition 1.** Let x\in\mathbb{Z} be non-zero. Then there exists a unique prime factorization x=\pm\prod_{i=1}^{s}{p_i^{e_i}}, where the p_i are distinct ascending primes and the e_i are positive integers. Then define, for each k\in\mathbb{Z}^+, the k-radical of x to be \text{rad}_k (x)=\prod_{i=1}^{s}{p_i^{\min(e_i,k)}}.

Note that for k=1, this is the classic radical function

\text{rad}_1 (x)=\prod_{i=1}^{s}{p_i^{\min(e_i,1)}=\prod_{i=1}^{s}{p_i}}=\text{rad}(x).

**Hu-Yang Conjecture.** Let \{a_i\}_{i=0}^{m} be integers with m\ge 2, such that \begin{cases} \gcd(a_0,a_1,\ldots, a_m) =1\\ a_1+\cdots+a_m=a_0\end{cases} and no subset of \{a_1,\ldots,a_m\} sums to 0. Then, independent of \{a_i\}_{i=0}^m,

\forall \epsilon>0,\exists \kappa >0 : \max_{0\le i\le m}(|a_i|)\le \kappa\cdot \left(\prod_{i=0}^m{\text{rad}_{(m-1)}(a_i)}\right)^{1+\epsilon}. \hspace{1cm} [\textbf{1}]

**Corollary.** The above conjecture would imply that for every \epsilon>0, for all but finitely many such \{a_i\}_{i=0}^m,

\max_{0\le i\le m}(|a_i|)\le \left(\prod_{i=0}^m{\text{rad}_{(m-1)}(a_i)}\right)^{1+\epsilon}.

**Lemma 1.** We have that

\forall x\in\mathbb{Z}, \forall n,k\in\mathbb{Z}^+: n\ge k\implies \text{rad}_k (x^n)=\text{rad}(x)^k.

*Proof.* Suppose x=\pm\prod_{i=1}^{s}{p_i^{e_i}} as in the definition, and that n,k\in\mathbb{Z}^+: n\ge k. Each e_i\ge 1 so ne_i\ge k and

\text{rad}_k (x^n)=\prod_{i=1}^{s}{p_i^{\min(ne_i,k)}}=\prod_{i=1}^{s}{p_i^k}=\left(\prod_{i=1}^{s}{p_i}\right)^k=\text{rad}(x)^k,

as desired.

**Proposition 1.** Assuming the conjecture:

- For all sufficiently large n, there exist no triples of positive

integers a\le b< c such that a^n+b^n=c^n+1. - If there are infinitely many triples of positive integers a\le b< c such that

a^n+b^n=c^n+1, then n\le 6.

*Proof.* Suppose that n,a,b,c\in\mathbb{Z}^+ such that a^n+b^n-c^n=1 with a\le b< c.

Then \gcd(a^n,b^n,-c^n,1)=1 and no subset of \{a^n,b^n,-c^n\} sums to 0. By the corollary and the lemma, for all but finitely many such triples (a,b,c),

c^n=\max(|a^n|,|b^n|,|-c^n|)\le \left(\text{rad}_2 (a^n)\cdot\text{rad}_2 (b^n)\cdot\text{rad}_2 (c^n)\right)^{1+\epsilon}= (a^2 b^2 c^2)^{1+\epsilon}<c^{6(1+\epsilon)}.

Fix any \epsilon>0. Then it is clear that for sufficiently large n, c does not even exist and so neither does a triple.

If the existence of infinitely many solutions is assumed, a triple with c^n<c^{6(1+\epsilon)} exists as there are only finitely many exceptions. Then n<6(1+\epsilon), and as this holds for every \epsilon>0, we get n\le 6.

**Corollary.** The exact same proposition and proof holds for the equation a^n+b^n=c^n -1.

In order to study the equations for even n, we first solve a^2+b^2=c^2\pm 1, for each sign.

**Lemma 2.** The binary quadratic form ap^2+bpq+cq^2 is equivalent to

- p^2+q^2 if and only if b^2-4ac=-4.
- p^2-q^2 if and only if b^2-4ac=4 and a,c are not both even.

**Proposition 2.** The integers x,y,z satisfy x^2+y^2=z^2\pm 1, for a fixed sign the opposite of which is \mp, if and only if there exist integers a_1,a_2,b_1,b_2 such that

\begin{cases}

1=a_1 b_2-a_2b_1\\

2x=(a_1^2-a_2^2)\mp (b_1^2-b_2^2)\\

y=a_1 a_2\mp b_1 b_2\\

2z=(a_1^2+a_2^2)\mp (b_1^2+b_2^2)

\end{cases}.

This is under the assumption that if not both of x,y are even, then x is odd, which does not compromise generality as x,y are symmetric in the equation.

*Proof.* Suppose x,y,z\in\mathbb{Z} such that x^2+y^2=z^2\pm 1, for a fixed sign. Let b=y and define a,c as

\begin{cases}a=z+x\\ c=z-x\end{cases}\iff \begin{cases}2x=a-c\\ 2z=a+c\end{cases}.

Substituting yields \left(\frac{a-c}{2}\right)^2+b^2=\left(\frac{a+c}{2}\right)^2\pm 1\iff b^2-ac=\pm 1.

Then the discriminant of the binary quadratic form \phi(p,q)=ap^2+2bpq+cq^2 is D(\phi)=(2b)^2-4ac=4(b^2-ac)=\pm 4. If D(\phi)=-4, then Lemma 2.1 can be immediately applied. If D(\phi)=4, then Lemma 2.2 applies. This is because at least one of a,c being odd is equivalent to at least one of z,x being odd. If z is odd, we are done. Otherwise, one of x,y must be odd and in the proposition we assumed that it is x in this case.

So ap^2+2bpq+cq^2 is equivalent to x^2\mp y^2. In other words, there exist a_1,a_2,b_1,b_2\in\mathbb{Z} such that

\begin{align*}

ap^2+2bpq+cq^2 &= (a_1 p+a_2 q)^2\mp (b_1 p+ b_2 q)^2\\

&= (a_1^2\mp b_1^2)p^2+2(a_1 a_2 \mp b_1 b_2)pq+(a_2^2\mp b_2^2)q^2.

\end{align*}

Moreover, since we have integers and each form is improperly equivalent to itself, it can be forced that

\begin{pmatrix}a_1&a_2\\ b_1& b_2\end{pmatrix}\in SL_2(\mathbb{Z})\implies\det\begin{pmatrix}a_1&a_2\\ b_1& b_2\end{pmatrix}=a_1 b_2 - a_2 b_1=1, One direction of the proposition follows from comparing coefficients. The other is straightforward by substituting and expanding.

So all solutions can be generated by SL_2(\mathbb{Z}).

**Lemma 3.** We have that a,b,c,d\in\mathbb{Z} satisfy

- a^2+b^2=c^2+d^2 if and only if there exist p,q,r,s\in\mathbb{Z} such that (a,b,c,d)=(pq+rs, pr-qs, pq-rs, pr+qs). This is assuming, without loss of generality, that a,c have the same parity and so do b,d.
- a^2+b^2+c^2=d^2 if and only if there exist p,q,r,s\in\mathbb{Z} such that (a,b,c,d)=(ps+qr, p^2-q^2+r^2-s^2, pq-rs, p^2+q^2+r^2+s^2).

*Proof.*

- Suppose a,b,c,d\in\mathbb{Z} satisfy a^2+b^2=c^2+d^2. Assume, without loss of generality, that a,c have the same parity and so do b,d. Define the integers (2a_0, 2b_0, 2c_0, 2d_0)=(a+c, d+b, a-c, d-b). so that the equation can be manipulated to a_0 c_0= b_0 d_0.

Since \mathbb{Z} is a unique factorization domain, there exist p,q,r,s\in\mathbb{Z} such that (a_0,b_0,c_0,d_0)=(pq,pr,rs,qs). Then (a,b,c,d)=(pq+rs,pr-qs,pq-rs,pr+qs).

The converse holds by Fibonacci's identity (pq+rs)^2+(pr-qs)^2=(pq-rs)^2+(pr+qs)^2.

- Suppose a,b,c,d\in\mathbb{Z} satisfy a^2+b^2+c^2=d^2. Rewrite the equation as (c+ai)(c-ai)=(d+b)(d-b) in Gaussian integers. Since \mathbb{Z}[i] is a unique factorization domain, there exist x=p+ri and y=q+si in \mathbb{Z}[i] such that (c+ai,c-ai, d+b, d-b)=(xy, \bar{x}\bar{y},x\bar{x}, y\bar{y}), where the bar denotes complex conjugation. Then \begin{cases} c+ai=(pq-rs) +(ps+qr)i\\ d+b=p^2+r^2\\ d-b=q^2+s^2\end{cases} so

(a,2b,c,2d)=(ps+qr, p^2-q^2+r^2-s^2, pq-rs, p^2+q^2+r^2+s^2).

The converse holds by Lebesgue's identity (2(ps+qr))^2+(p^2-q^2+r^2-s^2)^2+(2(pq-rs))^2=(p^2+q^2+r^2+s^2)^2.

**Attribution***Source : Link , Question Author : Dmitry Kamenetsky , Answer Author : Favst*