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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 Ali Caglayan
3 years 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 :)

Joel Tan - 3 years ago

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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

Subrata Saha - 3 years ago

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Why do you get a contradiction if p|y?

Bogdan Simeonov - 3 years ago

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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).

Tan Wee Kean - 3 years ago

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

Subrata Saha - 3 years ago

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Comment deleted Jan 28, 2015

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Good solution, @Ishan Singh

Anuj Shikarkhane - 2 years, 11 months ago

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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 :)

Ali Caglayan - 3 years ago

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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.

Calvin Lin Staff - 3 years ago

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

Ali Caglayan - 3 years ago

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?

Sanjay Sony - 2 years, 11 months ago

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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)

U Z - 2 years, 12 months ago

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

Ishan Singh - 2 years, 11 months ago

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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

U Z - 2 years, 11 months ago

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

Ali Caglayan - 2 years, 11 months ago

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@Ali Caglayan No, I don't think so.

Anuj Shikarkhane - 2 years, 11 months ago

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

Ishan Singh - 2 years, 11 months ago

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@Ishan Singh 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

U Z - 2 years, 11 months ago

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@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 \(\sqrt{34}\).

Ishan Singh - 2 years, 11 months ago

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@Ishan Singh Yes sorry I understood

U Z - 2 years, 11 months ago

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@Ali Caglayan

Kunal Jadhav - 2 years, 12 months ago

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