# Prove that there are no solutions to $x^2+1=y^3$

Prove that $x^2+1=y^3$ has no positive integer solutions.

Sure Catalan's conjecture works but can you do this with elementary number theory?

I ask about this on MSE here. Note that there is a valid argument using Gaussian integers and UFD's however I do not consider this elementary.

Problem status: Unsolved Note by A Former Brilliant Member
5 years, 3 months ago

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Firstly $x^{2}=y^{3}-1=(y-1)(y^{2}+y+1)$ Suppose $y-1$, $y^{2}+y+1$ were coprime. Then $y^{2}+y+1$ is a perfect square. Contradiction, since it is strictly between $y^{2}$ and $(y+1)^{2}$.

Suppose they weren't coprime. Then suppose a prime $p$ is a common divisor. Then $p|(y-1)^{2}=y^{2}-2y+1, p|y^{2}+y+1$ hence $p|3y$. If $p|y$, contradiction. Hence $p=3$. This means that their highest common factor is a power of 3. It can be checked (let y=1, 2, ..., 9) that $y^{2}+y+1$ is never a multiple of 9. Hence we know that their highest common factor is 3.

We know $x^{2}$ would be a multiple of 9. Hence $y-1=3m^{2}, y^{2}+y+1=3n^{2}$ for some coprime positive integers $m, n$. Consider the last equation. Expressing it as a quadratic equation in y, the discriminant is $12n^{2}-3$. This must be a perfect square. Let it be $9a^{2}$. Also let $b=2n$. Then $3b^{2}-3=9a^{2}, b^{2}-1=3a^{2}, b^{2}-3a^{2}=1$. This is a Pell's equation. We consider powers of $2+\sqrt 3$ as these will give us solutions.

We need $b$ to be even, so it must be an odd power. That's all I have so far :)

- 5 years, 3 months ago

Also, if you add the condition that y-1 is a perfect square, to (2+√3)^(2k+1). And manipulate with the coefficient of √3 , you get a contradiction

- 5 years, 3 months ago

Really awesome!! I was too thinking of this method. MIGHTY IS THE POWER OF FACTORISATION!!

- 5 years, 3 months ago

Why do you get a contradiction if p|y?

- 5 years, 3 months ago

P divides (y-1) to so it can't divide y unless p=1 (in which case the two numbers are coprime).

- 5 years, 3 months ago

A not-so-elementary solution:

Notice that $y$ must be odd since otherwise $y^3 \equiv 0 \pmod{4} \implies x^2 \equiv -1 \pmod{4},$ but $-1$ is not a quadratic residue modulo $4$. Now rearrange the equation as $(x+i)(x-i)= y^3$ (where $i^2=-1$) and consider this equation in Gaussian integers $\mathbb{Z}[i]$. Let $p$ be any (Gaussian) prime dividing $x+i$ and $x+i$. Then, $p$ must also divide $x+i-(x-i)= 2i$. The prime factorization of $2$ is $-i(i+1)^2,$ so $p \mid i^2(i+1)^2$. Hence, $p$ must be either $i$ or $i+1$. However, none of them are divisors of $x+i$ because all multiples of $i$ have real part $0$ and all multiples of $i+1$ are of the form $a+bi$ where $a$ and $b$ share the same parity. Hence, $x+i$ and $x-i$ are coprime, and since their product is a perfect cube, they must themselves be perfect cubes. Say $x+i= (a+ib)^3 = a^3 - 3ab^2 + (3a^2b-b^3)i$. Comparing the imaginary parts, we see that $3a^2b-b^3= 1,$ so $b \mid 1 \implies b = \pm 1$. Hence, either $3a^2-1 = 1$ or $-3a^2+1=1$. The former equation has no integer solutions; the latter gives $a=0$. For $(a, b)= (0, -1),$ we get $(a-ib)^3= i,$ so $x = 0$. Plugging this in the original equation, we get $y=1,$ so $(x, y)= (0, 1)$ is the only solution.

- 5 years, 3 months ago

Yes. Gaussian integers argument is beautiful for the problem :)

- 5 years, 3 months ago

Hm, the previous time I tried this using an argument involving factoring / divisibility, I ran into numerous difficulties. I would also be interested in an elementary proof.

Staff - 5 years, 3 months ago

Yes factorising divisibility arguments that I have seen so far all run into similar problems. Including Antonio's solution.

- 5 years, 3 months ago

$x^{2} + 1 = y^{3}$

$= x^{2} + 3^{2} = y^{3} + 2^{3}$ ( case of positive integers)

since L.H.S is a Pythagorean triplet , it will give a perfect square

and there is no perfect square which can be written as sum of 2 cubes , therefore no positive integers

now negative integers

lhs will remain positive and will give a perfect square and y = -1 is only possible as the negative integer . when y = -1 rhs = 7 which is not a perfect square

therefore no integral solutions are possible

Please point out my mistakes( I think I am wrong in the 4th line , please give example for it)

- 5 years, 3 months ago

- 5 years, 3 months ago

How is $x^2+3^2$ a Pythagorean Triplet?

- 5 years, 3 months ago

i don't know but $x^{2} + 3^{2}$ can be considered as a right angled triangle with perpendicular$=x^{2}$ and base as 9

- 5 years, 3 months ago

@U Z Will a sum of two squares always give a square number?

- 5 years, 3 months ago

No, I don't think so.

- 5 years, 3 months ago

@U Z But the hypotenuse need not be an integer. For example the triplet $(3,5,\sqrt{34})$.

- 5 years, 3 months ago

Oh yes thank you but let us suppose that x and y are integers then $y^{3} + 8$ should also be an integer and thus lhs should be a Pythagorean triplet and after the argument in my answer it can be proved that the condition we have taken is contradiction thus x and y are not integers

- 5 years, 3 months ago

@U Z Indeed $x$, $y$ and $y^3+8$ are integers. But this only proves that the square of the hypotenuse is an integer and not the hypotenuse itself. For instance in my previous example, $34$ (the square of the hypotenuse) is an integer but not $\sqrt{34}$.

- 5 years, 3 months ago

Yes sorry I understood

- 5 years, 3 months ago

?

- 5 years, 2 months ago