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Show that $x^3-x^2+8=y^2$ has no integer solution.

I spent several hours on this problem but I couldn't figure out how to solve it. A hint from the professor was to find a proper field where to find the solutions, or use Gaussian integers, but I still can't find the proof. I only showed that $y$ must be even, and I tried to factorize $(x^3+8)$ and equalize the two members with $x^2+y^2$, but nothing came out of it.

This problem was given at the Competencia Interuniversitaria Matemática Argentina (CIMA).

Rodrigo de Azevedo
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Michele Mascherpa
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  • Suppose that there are integer numbers $x$ and $y$ such that hold in the equation. The equation can be changed to $(x+2)(x^2-2x+4)=x^2+y^2$. Now, because of the equation $x^2-2x+4$ over integer numbers is ineducable, we can conclude that $$ x+2=\frac{x^2+y^2}{x^2-2x+4} $$ if $x$ be integer number then $x+2$ is integer number too and so, I suggest to think why the relation $\frac{x^2+y^2}{x^2-2x+4}$ has no integer solution? – Amin235 Jun 02 '17 at 11:02
  • i think somehow could be useful Fermat's theorem on sums of two squares but i don't know how to proceed – Michele Mascherpa Jun 02 '17 at 15:10

2 Answers2

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The factorization in the comment by Amin235 is the key. If $x$ is odd, then the term $x^2-2x+4$ is always $-1$ modulo $4$ (and hence has a prime divisor with the same property, which must divide $x^2+y^2$, a contradiction). If $x$ is even, then it is necessarily $2$ modulo $4$ (since otherwise $y^2$ is divisible by $8$, but not by $16$), whereby the same is true of $y$. It follows that $x^2+y^2$ is $8$ modulo $16$, while $(x+2)(x^2-2x+4)$ is $0$ modulo $16$.

Mike Bennett
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  • Needs more detail: if prime $q \equiv 3 \pmod 4$ and $q | (x^2 + y^2),$ then $q|x$ and $q|y$ so $q|\gcd(x,y)$ – Will Jagy Jun 02 '17 at 19:08
  • Compare my answer and comments yesterday, the student simply never got it: https://math.stackexchange.com/questions/2305956/solve-x3-y2-7/2306008#2306008 – Will Jagy Jun 02 '17 at 19:19
  • x^2(x-1)=[y-2^(3/2)][y+2^3/2] . How can x be integer where right side of equation is irrational? – sirous Jun 06 '17 at 08:11
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Here is an official solution: https://drive.google.com/file/d/1qOPYHK3p_S8tlwlewQ91X6JbDrZ-oKrc/view. Following is an English version (not literal translation, also little reordered):

If $x$ is even, say $x=2a$, we have $y^2=x^3-x^2+8=8a^3-4a^2+8$, implying $2 \mid y^2$, and so $2 \mid y$. Let $y=2b$, giving $8a^3-4a^2+8=4b^2$, and after simplification $2a^3-a^2+2=b^2$. Now if $a$ is even, then we have $b^2 \equiv 2 \pmod 4$, and if $a$ is odd, then $b^2 \equiv 3 \pmod 4$, either way impossible.

If $x \equiv 3 \pmod{4}$, then left side $x^3-x^2-8\equiv 2 \pmod{4}$. This implies $y^2 \equiv 2 \pmod {4}$, impossible.

For final case $x \equiv 1 \pmod {4}$, consider $$(x+2)(x^2-2x+4)=x^2+y^2.$$ Since we have $x+2 \equiv 3 \pmod 4$, there must be prime $p \equiv 3 \pmod {4}$ such that $p \mid x+2$. From equation above $p \mid x^2+y^2$, or $x^2 \equiv -y^2 \pmod {p}$. So $p \mid x$ iff $p \mid y$. We also have $$ x^{p-1} \equiv (x^2)^{\frac{p-1}{2}} \equiv (-y^2)^{\frac{p-1}{2}}\equiv (-1)^{\frac{p-1}{2}}y^{p-1} \pmod{p}. $$ Now clearly $p \nmid x$ (otherwise we would have $p=2$, a contradiction). So we have $p\nmid x, p \nmid y$. From Fermat's Little Theorem we have $x^{p-1}\equiv y^{p-1} \equiv 1 \pmod p$, and so from above equation $(-1)^{\frac{p-1}{2}}\equiv 1 \pmod{p}$. This implies $x\equiv 3 \pmod{p}$, a contradiction.

So there is no integer solution to the equation.

Sil
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