Prove that there are no solutions to x2+1=y3x^2+1=y^3

Prove that x2+1=y3x^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
4 years, 10 months ago

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Firstly x2=y31=(y1)(y2+y+1)x^{2}=y^{3}-1=(y-1)(y^{2}+y+1) Suppose y1y-1, y2+y+1y^{2}+y+1 were coprime. Then y2+y+1y^{2}+y+1 is a perfect square. Contradiction, since it is strictly between y2y^{2} and (y+1)2(y+1)^{2}.

Suppose they weren't coprime. Then suppose a prime pp is a common divisor. Then p(y1)2=y22y+1,py2+y+1p|(y-1)^{2}=y^{2}-2y+1, p|y^{2}+y+1 hence p3yp|3y. If pyp|y, contradiction. Hence p=3p=3. This means that their highest common factor is a power of 3. It can be checked (let y=1, 2, ..., 9) that y2+y+1y^{2}+y+1 is never a multiple of 9. Hence we know that their highest common factor is 3.

We know x2x^{2} would be a multiple of 9. Hence y1=3m2,y2+y+1=3n2y-1=3m^{2}, y^{2}+y+1=3n^{2} for some coprime positive integers m,nm, n. Consider the last equation. Expressing it as a quadratic equation in y, the discriminant is 12n2312n^{2}-3. This must be a perfect square. Let it be 9a29a^{2}. Also let b=2nb=2n. Then 3b23=9a2,b21=3a2,b23a2=13b^{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+32+\sqrt 3 as these will give us solutions.

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

Joel Tan - 4 years, 10 months 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

Brilliant Member - 4 years, 10 months ago

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

Brilliant Member - 4 years, 10 months ago

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Why do you get a contradiction if p|y?

Bogdan Simeonov - 4 years, 10 months 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 - 4 years, 10 months ago

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A not-so-elementary solution:

Notice that yy must be odd since otherwise y30(mod4)    x21(mod4),y^3 \equiv 0 \pmod{4} \implies x^2 \equiv -1 \pmod{4}, but 1-1 is not a quadratic residue modulo 44. Now rearrange the equation as (x+i)(xi)=y3(x+i)(x-i)= y^3 (where i2=1i^2=-1) and consider this equation in Gaussian integers Z[i]\mathbb{Z}[i]. Let pp be any (Gaussian) prime dividing x+ix+i and x+ix+i. Then, pp must also divide x+i(xi)=2ix+i-(x-i)= 2i. The prime factorization of 22 is i(i+1)2,-i(i+1)^2, so pi2(i+1)2p \mid i^2(i+1)^2. Hence, pp must be either ii or i+1i+1. However, none of them are divisors of x+ix+i because all multiples of ii have real part 00 and all multiples of i+1i+1 are of the form a+bia+bi where aa and bb share the same parity. Hence, x+ix+i and xix-i are coprime, and since their product is a perfect cube, they must themselves be perfect cubes. Say x+i=(a+ib)3=a33ab2+(3a2bb3)ix+i= (a+ib)^3 = a^3 - 3ab^2 + (3a^2b-b^3)i. Comparing the imaginary parts, we see that 3a2bb3=1,3a^2b-b^3= 1, so b1    b=±1b \mid 1 \implies b = \pm 1. Hence, either 3a21=13a^2-1 = 1 or 3a2+1=1-3a^2+1=1. The former equation has no integer solutions; the latter gives a=0a=0. For (a,b)=(0,1),(a, b)= (0, -1), we get (aib)3=i,(a-ib)^3= i, so x=0x = 0. Plugging this in the original equation, we get y=1,y=1, so (x,y)=(0,1)(x, y)= (0, 1) is the only solution.

Sreejato Bhattacharya - 4 years, 10 months ago

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

Ali Caglayan - 4 years, 10 months 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 - 4 years, 10 months 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 - 4 years, 10 months ago

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x2+1=y3x^{2} + 1 = y^{3}

=x2+32=y3+23 = 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 - 4 years, 9 months ago

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@Ali Caglayan

Kunal Jadhav - 4 years, 9 months ago

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How is x2+32x^2+3^2 a Pythagorean Triplet?

Ishan Singh - 4 years, 9 months ago

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i don't know but x2+32x^{2} + 3^{2} can be considered as a right angled triangle with perpendicular=x2=x^{2} and base as 9

U Z - 4 years, 9 months ago

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

Ali Caglayan - 4 years, 9 months ago

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

Anuj Shikarkhane - 4 years, 9 months ago

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

Ishan Singh - 4 years, 9 months ago

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@Ishan Singh Oh yes thank you but let us suppose that x and y are integers then y3+8y^{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 - 4 years, 9 months ago

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@U Z Indeed xx, yy and y3+8y^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, 3434 (the square of the hypotenuse) is an integer but not 34\sqrt{34}.

Ishan Singh - 4 years, 9 months ago

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@Ishan Singh Yes sorry I understood

U Z - 4 years, 9 months ago

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?

Sanjay Sony - 4 years, 9 months ago

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