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

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## Comments

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TopNewestFirstly \(x^{2}=y^{3}-1=(y-1)(y^{2}+y+1)\) 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 :)

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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

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Why do you get a contradiction if p|y?

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

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

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Comment deleted Jan 28, 2015

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Good solution, @Ishan Singh

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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.

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

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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.

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

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?

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\[x^{2} + 1 = y^{3}\]

\[ = x^{2} + 3^{2} = y^{3} + 2^{3}\] ( 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)

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How is \(x^2+3^2\) a Pythagorean Triplet?

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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

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squareof 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 \(\sqrt{34}\).Log in to reply

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@Ali Caglayan

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