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I would like to find all positive integer solutions for the below equation:

$1+n^3=m^4$

I suspect that it has no solution. How to prove that no solution exists?

I'm interested in knowing how to begin with solving such questions more than the particular solution of that question.

Is there a standard method to use (such like solving a quadratic equation) for these types of questions?

Hint on how to start solving the problem would also help.

I tried many algebraic manipulations like writing it as a difference of two squares twice, but didn't get to any where. I only figured that $n$ can't be a prime.

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  • $\begingroup$ A generalization of the problem: Find all positive integer solutions for the following equation: $l^2+n^3=m^4$ $\endgroup$
    – I0_0I
    Commented Oct 12, 2021 at 15:32
  • $\begingroup$ Yes I did program couldn't find a single one $\endgroup$
    – user960916
    Commented Oct 12, 2021 at 15:38
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    $\begingroup$ Modulo $9$ we have $0,1,2,5$ on the left $0,1,4,7$ on the right ... $\endgroup$ Commented Oct 12, 2021 at 15:43
  • $\begingroup$ Also you may refer to Mihăilescu's theorem that $x^{a}-y^{b} = 1$ implies that if $a,b > 1$ and $x,y > 0$, then; in particular, $a=2$, which is clearly not the case here. Hence, there are no natural solutions. $\endgroup$
    – Derek Luna
    Commented Oct 12, 2021 at 22:33

2 Answers 2

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Let $m$ and $n$ be integers such that $1+n^3=m^4$. Then also $$n^3=m^4-1=(m-1)(m+1)(m^2+1).\tag{1}$$ The greatest common divisor of any pair of factors on the right hand side divides $2$.

If $m$ is even then all factors are odd, and hence they pairwise coprime, and hence they are all perfect cubes. In particular $m-1$ and $m+1$ are perfect cubes. Of course the only two perfect cubes that differ by $2$ are $1$ and $-1$. This shows that $m=0$ and so $n=-1$.

On the other hand, if $m$ is odd then $n$ is even, then there are integers $a$ and $b$ such that $m=2a+1$ and $n=2b$. Plugging this into $(1)$ then leads to $$b^3=a(a+1)(2a^2+2a+1).$$ Again the three factors on the right hand side are pairwise coprime, and hence they are all perfect cubes. In particular $a$ and $a+1$ are perfect cubes. This means either $a=-1$ or $a=0$, corresponding to $m=-1$ and $m=1$, respectively, both yielding a solution with $n=0$.

This shows that the only integral solutions $(m,n)$ are $(-1,0)$, $(0,-1)$ and $(1,0)$. In particular, there are no solutions in the positive integers.

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    $\begingroup$ Brilliant solution! If I could accept 2 answers I would have accepted both answers to this question. $\endgroup$
    – I0_0I
    Commented Oct 12, 2021 at 19:45
  • $\begingroup$ +1. My edit was for a typo. When a=-1 or a=0 then n=0; not n=1. $\endgroup$ Commented Oct 12, 2021 at 21:29
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The more general Diophantine equation $$ X^2 + Y^3 = Z^4 \tag{1}$$ is an example of a generalized Fermat equation. The generic form of such an equation is $$ aX^p + bY^q = cZ^r. $$ There is a large literature on methods to study solutions to the generalized Fermat equation. In generality, see "The generalized Fermat equation" by Bennett, Mihailescu, and Siksek, (online here) or "The generalized Fermat equation" by Beukers (online here) or "On equations $z^m = F(x, y)$ and $Ax^p + By^q = Cz^r$ by Darmon and Granville (online here).

The Diophantine equation $(1)$ is studied completely in section 14.4 of Number Theory Volume II: Analytic and Modern Tools by Henri Cohen. Cohen parametrizes all solutions to $(1)$ in terms of seven families of solutions. To answer your question when $X = 1$, one could try to study the parametrized solutions. (I must admit that this itself is a nontrivial Diophantine exercise for some of the families, and I did not attempt to work this out fully).

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  • $\begingroup$ Thanks for the interesting references. (+1) $\endgroup$ Commented Oct 12, 2021 at 18:15

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