# Mathematical proof that $m^4+8$ is not a cube of an integer if $13$ does not divide $m$ (Table proof provided)

My question is how to express my solution to $m^4+8$ is not a cube of an integer if $13$ does not divide $m$ as a mathematical proof.

I understand that to prove by contradiction, the result is to be a cube of an integer:

\begin{align} m^4 + 8 & = n^3 \\ m^4 &= n^3 - 8 \\ \therefore m^4 &\equiv n^3-8\pmod{13}\\ \\ \end{align}

That equivalence is however impossible, because of the following table:

We see that $m^4 \equiv n^3-8\pmod{13}$ can only occur when m $\equiv 0 \pmod{13}$ and n = 2, 5 or 6.

This shows that since $13$ can not divide $m$, $m^4+8$ is not a cube of an integer. I am now sure how to express this mathematically.

Any help would be appreciated!

• Try to show that the expression (n - 2) [(n + 1)^2 + 3] is always even congruent to mod 13 and compare it with the result of your first table – saisanjeev Feb 9 '18 at 5:34
• What you have done is perfectly OK. Possibly there might be an easier way to do it, I don't know about this. – David Feb 9 '18 at 5:35
• To be honest, I think your proof is already pretty “mathematical”. It’s a proof by cases, which is common enough. – G Tony Jacobs Feb 9 '18 at 5:38

Your method is perfectly fine for this. Just for fun, I'm giving an alternative method in which you can make do with only a table of squares mod 13.

n     0  1  2  3  4  5  6  7  8  9 10 11 12
n^2   0  1  4  9  3 12 10 10 12  3  9  4  1


$$m^4+8 \equiv n^3 \mod 13\\ (m^4+8)^4 \equiv n^{12} \mod 13$$ From Fermat's little theorem we know that $n^{12}\equiv0,1 \mod 13$, so we then get: $$(m^4+8)^4 \equiv 0,1 \mod 13\\ (m^4+8)^2 \equiv 0,1,12 \mod 13\\ m^4+8 \equiv 0,1,5,8,12 \mod 13\\ m^4 \equiv 0,4,5,6,10 \mod 13$$ Note that $5$ and $6$ are not squares, so they are certainly not fourth powers. $$m^4 \equiv 0,4,10 \mod 13\\ m^2 \equiv 0,2,6,7,11 \mod 13$$ Note that the only one of these that is a square is $0$ so we are just left with $$m^2 \equiv 0 \mod 13\\ m \equiv 0 \mod 13$$

Consider the equation $n^3=m^2+8$. These types of equations are called Mordell equations.

Claim: The only integer solution to $n^3=m^2+8$ is given by $m=0, n=2$.

Suppose we have an integer solution to $n^3=m^2+8$. (All variables that are introduced will have integer values.)

Suppose $m$ is even, then $n$ must also be even, write $n=2x$ and $m=2m'$, then we get $8x'^3=4m'^2+8 \Rightarrow 2x'^3=m'^2+2$, so $m'$ is even, write $m'=2y$, we get $2x^3=4y^2+2$, so $x^3=2y^2+1$

To determine the solutions of $x^3=1+2y^2$, we will work in $\Bbb Z[\sqrt{-2}]$, this is known to be a PID.
$x^3=1+2y^2=(1+\sqrt{-2}y)(1-\sqrt{-2}y)$. Let $d=\gcd(1+\sqrt{-2}y,1-\sqrt{2}y)$ $d$ divides $(1+\sqrt{-2}y)+(1-\sqrt{-2}y)=2$. The prime factorization of $2$ in $\Bbb Z[\sqrt{-2}]$ is $2=- \sqrt{-2}^2$, so if $d \neq 1$, up to a sign $d$ is either $2$ or $\sqrt{-2}$, but then $2 \mid d^2 \mid (1+\sqrt{-2}y)(1-\sqrt{-2}y)=1+2y^2$, so $2 \mid 1 + 2y^2$, which is impossible because $\frac{1+2y^2}{2}=\frac{1}{2}+y^2 \not \in \Bbb Z[\sqrt{-2}]$, thus we get that $d=1$.

This means that $1+\sqrt{-2}y$ and $1-\sqrt{-2}y$ don't have any prime factor in common, but because their product is a third power, they must be themselves third powers. (Technically only up to a unit, but since the only units of $\Bbb Z[\sqrt{-2}]$ are $\pm 1$ and $(-1)^3=-1$, we can ignore this.)

Thus we get $$1+\sqrt{-2}y=(a+b\sqrt{-2})^3=a^3+3a^2b\sqrt{-2}-6ab^2-2b^3\sqrt{-2}=(a^3-6ab^2)+(3a^2b-2b^3)\sqrt{-2}$$ Now comparing coefficients gives $1=a^3-6ab^2$ and $y=3a^2b-2b^3$ From the first equation, we get $1=a(a^2-6b^2)$ so $a= \pm 1$. If $a=-1$, we must have $-1=a^2-6b^2=1-6b^2$, so $-2=-6b^2$, which has no solution. Thus $a=1$ and we get $1=a^2-6b^2=1-6b^2$, so $b=0$, thus means $1+\sqrt{-2}y=1$, so $y=0$. Thus $x^3=1+2y^2=1$, so $x=1$ Pluggin this back into our original equation gives $m=0$, $n=2$.

Now we come to the second case, where $m$ is odd, then $n$ must also be odd and we get $n^3=m^2+8=(m+2\sqrt{-2})(m-2\sqrt{-2})$
Let $d=\gcd(m+2\sqrt{-2},m-2\sqrt{-2})$, then $d$ divides $m+2\sqrt{-2} - (m-2\sqrt{-2})=4 \sqrt{-2}$. The prime factorization of $4\sqrt{-2}$ is $\sqrt{-2}^5$ So either $d=1$, or $\sqrt{-2} \mid d$. In the latter case, we get $-2 = d^2 \mid (m+2\sqrt{-2})(m-2\sqrt{-2})=m^2+8=n^3$, but this is impossible because $n$ is odd. Thus $d=1$ and $m+2\sqrt{-2}$ and $m-2\sqrt{-2}$ are coprime.
By the same argument as in the even case, we get $m+2\sqrt{-2}=(a+b\sqrt{-2})^3=(a^3-6ab^2)+(3a^2b-2b^3)\sqrt{-2}$
So comparing coefficients gives $3a^2b-2b^3=2$ and $m=a^3-6ab^2$, so $3a^2b=2(1+b^3)$, thus $2 \mid a$ or $2 \mid b$.
If $2 \mid a$, write $a=2a'$, we get $12a'^2b=2(1+b^3)$, so $6a'^2b=1+b^3$, thus $(6a'^2-b^2)b=1$, so $b = \pm 1$. If $b=1$, we get $1=6a'^2-b^2=6a'^2-1$, so $6a'^2=2$, which is impossible. If $b=-1$, we get $-1=6a'^2-b^2=6a'^2-1$, so $0=a'=a$, thus $m=a^3-6ab^2=0$, which is impossible, because we assumed that $m$ is odd.
If $2 \mid b$, write $b=2b'$, we get $6a^2b'-16b'^3=2$, so $(3a^2-8b'^2)b'=1$, thus $b'= \pm 1$. If $b'= -1$, we get $-1=3a^2-8b'^2=3a^2-8$, so $7=3a^2$ which has no solution. If $b'=1$, we get $1=3a^2-8b'^2=3a^2-8$, so $9=3a^2$, thus $3=a^2$, which has no solution. Thus there are no solutions where $m$ is odd.

To see how this related to the original question, note that if $n^3=m^4+8=(m^2)^2+8$, we can make the substituion $m \mapsto m^2$, to get a solution to $n^3=m^2+8$. But we know from the above discussion that the only such solution occurs when $m=0$ which also means $m=0$ in the equation $n^3=m^4+8$. In particular, there are no solutions where $13$ does not divide $m$.