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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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 :) – Joel Tan · 2 years, 3 months ago

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– Subrata Saha · 2 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 contradictionLog in to reply

– Bogdan Simeonov · 2 years, 3 months ago

Why do you get a contradiction if p|y?Log in to reply

– Tan Wee Kean · 2 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).Log in to reply

– Subrata Saha · 2 years, 3 months ago

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

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@Ishan Singh – Anuj Shikarkhane · 2 years, 2 months ago

Good solution,Log in to reply

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. – Sreejato Bhattacharya · 2 years, 3 months ago

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– Ali Caglayan · 2 years, 3 months ago

Yes. Gaussian integers argument is beautiful for the problem :)Log in to reply

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 · 2 years, 3 months ago

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– Ali Caglayan · 2 years, 3 months ago

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

? – Sanjay Sony · 2 years, 2 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) – Megh Choksi · 2 years, 2 months ago

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– Ishan Singh · 2 years, 2 months ago

How is \(x^2+3^2\) a Pythagorean Triplet?Log in to reply

– Megh Choksi · 2 years, 2 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 9Log in to reply

– Ali Caglayan · 2 years, 2 months ago

Will a sum of two squares always give a square number?Log in to reply

– Anuj Shikarkhane · 2 years, 2 months ago

No, I don't think so.Log in to reply

– Ishan Singh · 2 years, 2 months ago

But the hypotenuse need not be an integer. For example the triplet \((3,5,\sqrt{34})\).Log in to reply

– Megh Choksi · 2 years, 2 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 integersLog in to reply

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 \(\sqrt{34}\). – Ishan Singh · 2 years, 2 months agoLog in to reply

– Megh Choksi · 2 years, 2 months ago

Yes sorry I understoodLog in to reply

@Ali Caglayan – Kunal Jadhav · 2 years, 2 months ago

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