# Challenge Math Challenge - August 2019

• Featured

#### nuuskur

Suppose $\alpha$ is $n$-th root of unity and $\alpha \in \mathbb F_{q^m}$ for some $m\in\mathbb N$. Then by Lagrange $n \mid q^m-1$. It would suffice to compute the order of $q$ modulo $n$. As $(q,n)=1$ it holds that $n\mid q^{\phi (n)} -1$, where $\phi$ is the Euler totient map. So the order of $q$ must be a divisor of $\phi (n)$.

Not sure how to explicitly compute $m$.

#### Math_QED

Homework Helper
Suppose $\alpha$ is $n$-th root of unity and $\alpha \in \mathbb F_{q^m}$ for some $m\in\mathbb N$. Then by Lagrange $n \mid q^m-1$. It would suffice to compute the order of $q$ modulo $n$. As $(q,n)=1$ it holds that $n\mid q^{\phi (n)} -1$, where $\phi$ is the Euler totient map. So the order of $q$ must be a divisor of $\phi (n)$.

Not sure how to explicitly compute $m$.
Try to take $m$ minimal somehow.

#### nuuskur

Try to take $m$ minimal somehow.
Uhh, this is like trial and error. It's about reducing the powers of $x:=\phi (n) = \prod _{i\in I}p_i^{k_i}$. I can think of recursion.

I'll write the most barebones version I can think of, can likely be (heavily) optimised. Let $P(x) \Leftrightarrow n\mid q^{x}-1$.

1. Fix $a\in I$.
2. If $\neg P(x/p_a)$, then return $x$ and terminate.
3. Else put $x:= x/p_a$ and go to 1 (with $I$-many branches)

Do this for every $a\in I$ as a starting point (there will be a lot of branching). Eventually, pick the smallest $x$ that was returned, this will be the order of $q$ modulo $n$.

Last edited:

#### Math_QED

Homework Helper
Try to take $m$ minimal somehow.
I'm not looking for a closed, explicit form of the order of q mod n. Just explain why it is finite.

#### nuuskur

I'm not looking for a closed, explicit form of the order of q mod n. Just explain why it is finite.
Well, per assumption $n\mid q^{\phi (n)}-1$. Thus the set $\{x : n\mid q^x -1\}\subset \mathbb N$ is non-empty, therefore there is smallest one.

#### Math_QED

Homework Helper
Well, per assumption $n\mid q^{\phi (n)}-1$. Thus the set $\{x : n\mid q^x -1\}\subset \mathbb N$ is non-empty, therefore there is smallest one.
Where did you (implicitely) use that $(q,n)=1$? And that does not solve the problem entirely of course.

#### nuuskur

Where did you (implicitely) use that $(q,n)=1$? And that does not solve the problem entirely of course.
$(q,n)=1$ guarantees $n\mid q^{\phi (n)}-1$ (Euler's theorem). It's not clear to me what solution is expected. I gave a way of finding the order $m$ and justified its existence. The smallest field extension with desired property would have to be $\mathbb F_{q^m}$.

Ok, maybe there's the issue of necessarily containing a subgroup of order $n$. But the multiplicative group of a finite field is cyclic, therefore there is a subgroup of any order that divides the order of the group.

Last edited:

#### Math_QED

Homework Helper
$(q,n)=1$ guarantees $n\mid q^{\phi (n)}-1$ (Euler's theorem). It's not clear to me what solution is expected. I gave a way of finding the order $m$ and justified its existence. The smallest field extension with desired property would have to be $\mathbb F_{q^m}$.

Ok, maybe there's the issue of necessarily containing a subgroup of order $n$. But the multiplicative group of a finite field is cyclic, therefore there is a subgroup of any order that divides the order of the group.
Yes, your first solution assumed there was an n-th root and then deduced a necessary condition. Your last edit shows that this necessary condition is also sufficient to have an n-th root.

For me, you solved the question succesfully!

#### nuuskur

Oh, no more problems remaining Well, that's it for me, folks. Likely not taking part in next month's competition. *drops the mic*

#### Math_QED

Homework Helper
Oh, no more problems remaining Well, that's it for me, folks. Likely not taking part in next month's competition. *drops the mic*
Other challenge threads still have open problems.

#### QuantumQuest

Gold Member
2. Find the equation of a curve such that $y′′$ is always $2$ and the slope of the tangent line is $10$ at the point $(2,6)$.

undefined
The way you solve it is correct but there are some mistakes - most likely typos, in the solution. In any case it is already solved by @KnotTheorist.

#### QuantumQuest

Gold Member
. The maximum value of $f$ with $f(x) = x^a e^{2a - x}$ is minimal for which values of positive numbers $a$ ?

undefined
Your way of thinking is correct. The question has already been solved by @Pi-is-3.

#### Kyle Nemeth

My attempt at 14 (I'm a novice so bare with me!)
Consider a solution $(a,b,c)$ to the equation $$2x^6+3y^6=z^3$$ where $a,b,c \neq 0$ and $a,b,c \in \mathbb{Q}$. This yields $$2a^6+3b^6=c^3$$ $$\implies 2\frac {a^6}{c^3}+3 \frac{b^6}{c^3}=1$$ $$\implies 2 \left( \frac {a^2}{c} \right)^3 +3 \left( \frac {b^2}{c} \right)^3=1$$ We search for the possibility of a rational solution. If $$\left( \frac {a^2}{c} \right)^3 \neq \frac {r^3}{t^3}$$ for $r,t \in \mathbb {Z}$, then $\sqrt[3] \frac {r}{t} \in \mathbb {I}$ $\forall r,t$ with no common factor, due to the prime factorization theorem. This would imply that either $a \in \mathbb {I}$ or $c \in \mathbb {I}$. The same logic follows for $\left( \frac {b^2}{c} \right)^3$. So for a rational solution to exist, it must be that $$\frac {a^2}{c} = \frac {r}{t}, \frac {b^2}{c} = \frac {s}{t}$$ where $s \in \mathbb {Z}$, and we have reduced $\frac{r}{t}$ and $\frac{s}{t}$ to simplest form, as can be done with any rational number. Then $$2 \left( \frac{r^3}{t^3} \right)+3 \left( \frac {s^3}{t^3} \right) = 1$$ $$\implies 2r^3 + 3s^3=t^3$$ With algebra, we can show that $$f(s,t)=\sqrt[3] \frac{t^3-3s^3}{2}$$ If $t,s$ are both even, then $\frac {s}{t}$ is not fully simplified, but we assumed that we did simplify $\frac {s}{t}$ previously, so this cannot be the case. If $t,s$ are even and odd respectively, or vice versa, then the numerator $t^3-3s^3$ is odd. By definition, an odd number does not contain a factor of 2 and so the numerator and denominator of $f(s,t)$ will be uniquely irrational (irrational with no common factor) and so $r$ will be irrational in this case, implying $a\in \mathbb{I}$ or $c\in \mathbb{I}$. If $s,t$ are both odd, then $t^3-3s^3$ is even and the numerator may contain a factor of 2. By the prime factorization theorem, we may write $$t^3-3s^3=2^k(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$$ where $k,n_i \in \mathbb {N}$ and $P_i$ is a prime number. Here, $2^k$ can be a perfect cube, but how do we know that $(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$ is never a perfect cube (which may imply a possible rational solution)? Consider the case where $s=t$. Then $$r = f(s,t)=\sqrt[3] \frac {t^3-3t^3}{2}=\sqrt[3] \frac {-2t^3}{2}=-\sqrt[3] {t^3}$$ $$\implies r^3=-t^3$$ From this we see that in the case where $s=t$, $(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$ must be a perfect cube, since $2r^3=t^3-3s^3$ (and it must be odd as well, since any factor of 2 from the prime factorization is contained within $2^k$). However, if $r=-t$ then we get a complex set of solutions from $\frac {a^2}{c}$ and $\frac {b^2}{c}$. It can be shown for the equation $2r^3+3s^3=t^3$ that there is a 1-1 correspondence over the rationals for three separate cases: $$1) r = f(s,t)$$ $$2) s=g(r,t)$$ $$3) t=h(s,r)$$ Thus, when $s \neq t$, it cannot be true that $(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$ is a perfect cube, which implies that in this case $r\in \mathbb{I}$ regardless of whether $2^k$ is a perfect cube or not. This would then imply that $a \in \mathbb{I}$ or $c \in \mathbb{I}$. Therefore, for any $s,t$, there cannot exist a possible rational solution. Similar cases can be proven with $g(r,t)$ and $h(s,r)$ based on the 1-1 correspondence. There only exists a trivial rational solution, $x=y=z=0$ since $$2(0)^6+3(0)^6=(0)^3$$ $$\implies 0=0$$

#### fresh_42

Mentor
2018 Award
My attempt at 14 (I'm a novice so bare with me!)
Sure. You put much effort in it, I appreciate that, and it is not an easy problem. Nevertheless I'll have to be strict.
Consider a solution $(a,b,c)$ to the equation $$2x^6+3y^6=z^3$$ where $a,b,c \neq 0$ and $a,b,c \in \mathbb{Q}$.
And what in the other cases? We may assume that not all three of them, or even two equal zero, since then we are done. But what is in the cases $c=0$ or $a=0$?
This yields $$2a^6+3b^6=c^3$$ $$\implies 2\frac {a^6}{c^3}+3 \frac{b^6}{c^3}=1$$ $$\implies 2 \left( \frac {a^2}{c} \right)^3 +3 \left( \frac {b^2}{c} \right)^3=1$$ We search for the possibility of a rational solution.
Now I get lost. What do you need $r,s,t$ for?
If $$\left( \frac {a^2}{c} \right)^3 \neq \frac {r^3}{t^3}$$ for $r,t \in \mathbb {Z}$, ...
This is never true. We can always write a quotient of two quotients as one quotient.
... then $\sqrt[3] \frac {r}{t} \in \mathbb {I}$ $\forall r,t$ with no common factor, due to the prime factorization theorem. This would imply that either $a \in \mathbb {I}$ or $c \in \mathbb {I}$. The same logic follows for $\left( \frac {b^2}{c} \right)^3$. So for a rational solution to exist, it must be that $$\frac {a^2}{c} = \frac {r}{t}, \frac {b^2}{c} = \frac {s}{t}$$ where $s \in \mathbb {Z}$, and we have reduced $\frac{r}{t}$ and $\frac{s}{t}$ to simplest form, as can be done with any rational number. Then $$2 \left( \frac{r^3}{t^3} \right)+3 \left( \frac {s^3}{t^3} \right) = 1$$ $$\implies 2r^3 + 3s^3=t^3$$
Now we have a solution $r=-1, s=t=1$ but no solution for the original question!
With algebra, we can show that $$f(s,t)=\sqrt[3] \frac{t^3-3s^3}{2}$$ If $t,s$ are both even, then $\frac {s}{t}$ is not fully simplified, but we assumed that we did simplify $\frac {s}{t}$ previously, so this cannot be the case. If $t,s$ are even and odd respectively, or vice versa, then the numerator $t^3-3s^3$ is odd. By definition, an odd number does not contain a factor of 2 and so the numerator and denominator of $f(s,t)$ will be uniquely irrational (irrational with no common factor) and so $r$ will be irrational in this case, implying $a\in \mathbb{I}$ or $c\in \mathbb{I}$. If $s,t$ are both odd, then $t^3-3s^3$ is even and the numerator may contain a factor of 2. By the prime factorization theorem, we may write $$t^3-3s^3=2^k(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$$ where $k,n_i \in \mathbb {N}$ and $P_i$ is a prime number. Here, $2^k$ can be a perfect cube, but how do we know that $(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$ is never a perfect cube (which may imply a possible rational solution)? Consider the case where $s=t$. Then $$r = f(s,t)=\sqrt[3] \frac {t^3-3t^3}{2}=\sqrt[3] \frac {-2t^3}{2}=-\sqrt[3] {t^3}$$ $$\implies r^3=-t^3$$ From this we see that in the case where $s=t$, $(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$ must be a perfect cube, since $2r^3=t^3-3s^3$ (and it must be odd as well, since any factor of 2 from the prime factorization is contained within $2^k$). However, if $r=-t$ then we get a complex set of solutions from $\frac {a^2}{c}$ and $\frac {b^2}{c}$. It can be shown ...
Never, ever use this platitude! You bet it is in nine of ten cases wrong and thus needlessly embarrassing.
You were on the right track. Divisibility is the key. You should consider the remainders of a division by $7$. There are not many remainders possible for cubes.

#### Kyle Nemeth

And what in the other cases? We may assume that not all three of them, or even two equal zero, since then we are done. But what is in the cases $c=0$ or $a=0$?
Understood. I am quite unsure exactly of the nature of a "rational solution," whether each $a,b,c$ have to all be rational or if a rational solution could be constituted by something like $(a,b,c)$ where only $a,b \in \mathbb{Q}$ and maybe $c\in \mathbb{R}$ for example. Either way, I understand that 0 is indeed rational, and so I have left out those cases.
Now I get lost. What do you need $r,s,t$ for?
My goal was to reduce the problem down to integers instead of working with rational numbers, since then I could apply the parity arguments.
This is never true. We can always write a quotient of two quotients as one quotient.
I believe that was actually the intention of my argument. If $a,c$ are rational numbers (each an integer divided by an integer), then $\frac{a^2}{c}$ can be written as a single fraction in reduced form, $\frac{r}{t}$ where $r,t$ are integers.
Now we have a solution $r=-1, s=t=1$ but no solution for the original question!
This is a solution, but it would be a complex solution to the original equation and so I was under the presumption that a solution of this form would not be accepted as a rational solution to the original question.
Never, ever use this platitude! You bet it is in nine of ten cases wrong and thus needlessly embarrassing.
Would you mind pointing out exactly where my logic fails within this argument? I value this problem as a learning experience as well as a challenge. Thank you for the haste reply!

#### fresh_42

Mentor
2018 Award
Most of what you wrote can be directly achieved. If we have an equation with quotients, we can always make it an integer equation, just multiply the quotients to get the least common multiple in the denominator.
Using the definition of a prime, namely $p$ is prime, if it is not $\pm 1$ and $p\,|\,ab$ requires $p\,|\,a$ or $p\,|\,b$ makes the cancellations you need. But the reduction to $2r^3+3s^3=t^3$ isn't the solution, since $(r,s,t)=(-1,1,1)$ is a counterexample.

I read your proof very carefully and achieved your conclusions with slightly different arguments. So maybe the proof can be corrected by the additional assumption, that all $r,s,t$ are positive. I got at least as far as $2r^3+3s^3=t^3$ with coprime pairs $(r,t)$ and $(s,t)$. But then the solution with $r=-1$ came to mind.

You could really try to pass all equations to their remainders modulo $7$ and look for common divisors.
If $2a^6+3b^6=c^3$ then the same equation holds if we substitute $a,b,c$ by their remainders of any division, esp. $7$.

An example of what I mean: $13^2=12^2+5^2$. Division by $5$ yields as remainders $9=3^2\equiv 2^2+0^2 = 4 \mod 5$, since $9$ and $4$ have the same remainder modulo a division by $5$. This works for any number. The above problem can be tackled by remainders modulo $7$. The remainders modulo seven are $\{\,0,1,2,3,4,5,6\,\}$ which is cubed $\{\,0,1,8,27,64,125,216\,\}$ with remainders $\{\,0,1,1,6,1,6,6\,\}=\{\,0,1,6\,\}$. This is a very limited selection of possibilities.

#### Kyle Nemeth

Understood, the substitutions I made were clearly not necessary and the fact that any equation with quotients can be written as an integer equation by multiplying the least common denominator is an approach I did not think of.
But the reduction to $2r^3+3s^3=t^3$ isn't the solution, since $(r,s,t)=(−1,1,1)$ is a counterexample.
I do not assert that this equation has no solutions, but rather any potential solution will ultimately give a solution to the original equation $$2x^6+3y^6=z^3$$ that is not rational.
So maybe the proof can be corrected by the additional assumption, that all $r,s,t$ are positive.
In order for $2x^6+3y^6=z^3$ to have a solution that is rational, isn't there a certain restriction on which values may be negative in a solution $(r,s,t)$? Since $$\frac {a^2}{c} = \frac {r}{t}, \frac {b^2}{c} = \frac {s}{t}$$ In order that $a$ or $b$ not be complex, for $c>0$, mustn't it be the case that a potential solution must exist of the forms $(r,s,t)$ where $r,s,t<0$ or $r,s,t>0$? or if $c<0$, then we must have $r,s<0$,$t>0$ or $r,s>0$, $t<0$? But if $r,s<0$ and $t>0$, we can let $r=-\alpha$, $s=-\beta$ and $t=\gamma$, for integers $\alpha,\beta,\gamma>0$, then (using $f(s,t)$ in post 113) $$-\alpha = \sqrt[3] \frac{\gamma^3-3(-\beta)^3}{2}$$ $$\implies -\alpha=\sqrt[3] \frac{\gamma^3+3\beta^3}{2}$$ and this equation is never true, since $\alpha,\beta,\gamma>0$, and this is similarly the case for where $r,s>0$, $t<0$.
And what in the other cases? We may assume that not all three of them, or even two equal zero, since then we are done. But what is in the cases $c=0$ or $a=0$?
If $a=0$, then $$3b^6=c^3$$ $$\implies \frac {c}{b^2}=\sqrt[3] 3$$ which implies that $c$ or $b$ must be irrational, since a number $\frac {p}{q}\in \mathbb{I}$ if either $p$ or $q$ is irrational. If $b=0$, then $$2a^6=c^3$$ $$\implies \frac{c}{a^2}=\sqrt[3] 2$$ and so $c$ or $a$ must be irrational. If $c=0$, then $$2a^6+3b^6=0$$ $$\implies \frac {a}{b}=\sqrt[6] \frac {-3}{2}$$ which also implies $a$ or $b$ must be irrational (and complex).
You could really try to pass all equations to their remainders modulo $7$ and look for common divisors.
Ahh I see, I currently do not have much knowledge of modular arithmetic so that is another approach I was not aware of and I can see how this approach is far more simple than mine.

#### fresh_42

Mentor
2018 Award
I will check later. I'm too tired now to say something reasonable, or to read something correctly.

We can assume $a.b,c > 0$ from the start, after the cases $c=0$ (easy) and $a=0$ (prime definition with $p=2$ if $b=0$ or symmetrically $p=3$ if $a=0$) are done. A solution with a negative number immediately also gets a solution with a positive number, and $c<0$ is impossible.
This leads to $2r^3+3s^3=t^3$ with positive integers $r,s,t$. That they may be assumed pairwise coprime is again a consequence of primality.

E.g. let us assume a prime $p$ divides $r$ and $t$. Then
\begin{align*}
p \,|\, 3s^3=t^3-2r^3 & \Longrightarrow p\,|\,3 \text{ or } p\,|\,s^3\\
\text{ (a) }p\,|\,s^3 &\Longrightarrow p\,|\,s\\
&\Longrightarrow p^3\,|\,s^3\\
&\Longrightarrow p^3 \text{ divides every term and can be cancelled }\\
\text{ (b) }p\,|\,3&\Longrightarrow p=3\\
&\Longrightarrow s^3= 9\cdot t'^3 - 2\cdot 9\cdot r'^3\\
&\Longrightarrow 3\,|\,s^3\\
&\Longrightarrow 3\,|\,s\\
&\Longrightarrow 3^3\,|\,s^3\\
&\Longrightarrow 3^3 \text{ divides every term and can be cancelled }
\end{align*}
This is the way primality is normally used. Here it allows us to assume that $r,s,t$ are pairwise coprime, since otherwise we would get a common factor cubed, which we can cancel and still have the same equation structure $2r'^3+3s'^3=t'^3$. So we continue to cancel all common factors. Now we have
• $2r^3+3s^3=t^3$
• $r,s,t > 0$
• $r,s,t$ are pairwise coprime
If $c=0$, then $2a^6+3b^6=0 \quad \implies \frac {a}{b}=\sqrt[6] \frac {-3}{2}$ which also implies $a$ or $b$ must be irrational (and complex).
You need $b\neq 0$ here, but this is not necessary. Both terms of the right hand side are always non negative. If their sum equals zero, then both summands have to be zero, i.e. $a=b=c=0$.

Last edited:

#### fresh_42

Mentor
2018 Award
Consider a solution $(a,b,c)$ to the equation
$$2x^6+3y^6=z^3$$
where $a,b,c \neq 0$ and $a,b,c \in \mathbb{Q}$. This yields
$$2a^6+3b^6=c^3$$
$$\implies 2\frac {a^6}{c^3}+3 \frac{b^6}{c^3}=1$$
$$\implies 2 \left( \frac {a^2}{c} \right)^3 +3 \left( \frac {b^2}{c} \right)^3=1$$
I skip the part here that is a bit confusing and unnecessary. We can always transform $2a^6+3b^6=c^3$ into ...
$$\implies 2r^3 + 3s^3=t^3$$
where we may assume w.l.o.g. $r,s,t > 0$ and $r,s,t$ are pairwise coprime integers (cp. post #118).
With algebra, we can show that
$$f(s,t)=\sqrt[3] \frac{t^3-3s^3}{2}$$
If $t,s$ are both even...
Obsolete, if we first guarantee that $s$ and $t$ are coprime.
... then $\frac {s}{t}$ is not fully simplified, but we assumed that we did simplify $\frac {s}{t}$ previously, so this cannot be the case. If $t,s$ are even and odd respectively, or vice versa, then the numerator $t^3-3s^3$ is odd. By definition, an odd number does not contain a factor of 2 and so the numerator and denominator of $f(s,t)$ will be uniquely irrational (irrational with no common factor) and so $r$ will be irrational in this case, implying $a\in \mathbb{I}$ or $c\in \mathbb{I}$.
What we really have in this case is, that a number $z+\frac{1}{2}$ is under the root. We need to show that there is no integer $r$ with $r^3=z+\frac{1}{2}$. But this is true and no argument involving irrationality is needed.
Keep it simple and only use what you have, step by step.
If $s,t$ are both odd, then $t^3-3s^3$ is even and the numerator may contain a factor of 2. By the prime factorization theorem, we may write
$$t^3-3s^3=2^k(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$$
where $k,n_i \in \mathbb {N}$ and $P_i$ is a prime number. Here, $2^k$ can be a perfect cube, but how do we know that $(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$ is never a perfect cube (which may imply a possible rational solution)? Consider the case where $s=t$. Then
$$r = f(s,t)=\sqrt[3] \frac {t^3-3t^3}{2}=\sqrt[3] \frac {-2t^3}{2}=-\sqrt[3] {t^3}$$
$$\implies r^3=-t^3$$
From this we see that in the case where $s=t$...
... which again is obsolete, since $s$ and $t$ can be assumed coprime.
...$(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$ must be a perfect cube, since $2r^3=t^3-3s^3$ (and it must be odd as well, since any factor of 2 from the prime factorization is contained within $2^k$). However, if $r=-t$ then we get a complex set of solutions from $\frac {a^2}{c}$ and $\frac {b^2}{c}$. It can be shown for the equation $2r^3+3s^3=t^3$ that there is a 1-1 correspondence over the rationals for three separate cases: $$1) r = f(s,t)$$
$$2) s=g(r,t)$$
$$3) t=h(s,r)$$
Summary: We are left with $s\neq t$, both odd, and
$$\dfrac{t^3-3s^3}{2} = \left( 2^k P_1^{n_1}\cdot P_2^{n_2}\cdot \ldots \cdot P_{n_k}^{n_k}\right)^{1/3} \in \mathbb{Z} \;\;\text{ i.e. }\;\; 3\,|\, k,n_j$$
Thus, when $s \neq t$, it cannot be true that $(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$ is a perfect cube...
Sorry, but I do not see this. Why is it impossible that all powers are divisible by three? What happens if they were? E.g. all zero is a possibility.
...which implies that in this case $r\in \mathbb{I}$ regardless of whether $2^k$ is a perfect cube or not. This would then imply that $a \in \mathbb{I}$ or $c \in \mathbb{I}$. Therefore, for any $s,t$, there cannot exist a possible rational solution. Similar cases can be proven with $g(r,t)$ and $h(s,r)$ based on the 1-1 correspondence. There only exists a trivial rational solution, $x=y=z=0$ since
$$2(0)^6+3(0)^6=(0)^3$$
$$\implies 0=0$$

Last edited:

#### Pi-is-3

You are hard to follow.

Done in post #89, although the additional variables make it unnecessarily complicated (see post #90).

We do not "see" it. What we see is that you invented out of the blue a certain number $t$. What you should have done is telling us, that you set $t:= \sqrt[3]{\frac{3}{2}y} \in \mathbb{R}$ such that $y=\frac{2}{3}t^3$. But now $x^3\neq \frac{1}{2}-t^3$.

To correct this, we have to set $t:=\sqrt[3]{\frac{3}{2}}y$. Now $x^3= \frac{1}{2}-t^3$.

How that?
Sorry for late reply. I was not able to come online a lot since a long time. Anyway, I tried to prove it, but I couldn't. Hence my proof is wrong. Sorry.

"Math Challenge - August 2019"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving