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
5 years, 1 month ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

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 - 5 years, 1 month ago

Log in to reply

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

A Former Brilliant Member - 5 years, 1 month ago

Log in to reply

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

A Former Brilliant Member - 5 years, 1 month ago

Log in to reply

Why do you get a contradiction if p|y?

Bogdan Simeonov - 5 years, 1 month ago

Log in to reply

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 - 5 years, 1 month ago

Log in to reply

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 - 5 years, 1 month ago

Log in to reply

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

Ali Caglayan - 5 years, 1 month ago

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 - 5 years, 1 month ago

Log in to reply

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

Ali Caglayan - 5 years, 1 month ago

Log in to reply

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 - 5 years ago

Log in to reply

Log in to reply

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

Ishan Singh - 5 years ago

Log in to reply

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 - 5 years ago

Log in to reply

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

Ali Caglayan - 5 years ago

Log in to reply

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

Anuj Shikarkhane - 5 years ago

Log in to reply

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

Ishan Singh - 5 years ago

Log in to reply

@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 - 5 years ago

Log in to reply

@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 - 5 years ago

Log in to reply

@Ishan Singh Yes sorry I understood

U Z - 5 years ago

Log in to reply

?

Sanjay Sony - 5 years ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...