- #101

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Not sure how to explicitly compute [itex]m[/itex].

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- #101

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Not sure how to explicitly compute [itex]m[/itex].

- #102

Math_QED

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Try to take ##m## minimal somehow.

Not sure how to explicitly compute [itex]m[/itex].

- #103

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

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

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

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

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- #104

Math_QED

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

- #105

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

- #106

Math_QED

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

- #107

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

Ok, maybe there's the issue of necessarily containing a subgroup of order [itex]n[/itex]. 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.

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- #108

Math_QED

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

Ok, maybe there's the issue of necessarily containing a subgroup of order [itex]n[/itex]. 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.

For me, you solved the question succesfully!

- #109

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- #110

Math_QED

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Other challenge threads still have open problems.*drops the mic*

- #111

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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.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)##.

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- #112

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Your way of thinking is correct. The question has already been solved by @Pi-is-3..The maximum value of ##f## with ##f(x) = x^a e^{2a - x}## is minimal for which values of positive numbers ##a## ?

undefined

- #113

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- #114

fresh_42

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Sure. You put much effort in it, I appreciate that, and it is not an easy problem. Nevertheless I'll have to be strict.My attempt at 14 (I'm a novice so bare with me!)

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##?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}##.

Now I get lost. What do you need ##r,s,t## for?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.

This is never true. We can always write a quotient of two quotients as one quotient.If $$ \left( \frac {a^2}{c} \right)^3 \neq \frac {r^3}{t^3}$$ for ##r,t \in \mathbb {Z}##, ...

Now we have a solution ##r=-1, s=t=1## but no solution for the original question!... 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 $$2 \left( \frac{r^3}{t^3} \right)+3 \left( \frac {s^3}{t^3} \right) = 1$$ $$\implies 2r^3 + 3s^3=t^3$$

Never, ever use this platitude! You bet it is in nine of ten cases wrong and thus needlessly embarrassing.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 ...

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.

- #115

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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.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##?

My goal was to reduce the problem down to integers instead of working with rational numbers, since then I could apply the parity arguments.Now I get lost. What do you need ##r,s,t## for?

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.This is never true. We can always write a quotient of two quotients as one quotient.

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.Now we have a solution ##r=-1, s=t=1## but no solution for the original question!

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!Never, ever use this platitude! You bet it is in nine of ten cases wrong and thus needlessly embarrassing.

- #116

fresh_42

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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.

- #117

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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.But the reduction to ##2r^3+3s^3=t^3## isn't the solution, since ##(r,s,t)=(−1,1,1)## is a counterexample.

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##.So maybe the proof can be corrected by the additional assumption, that all ##r,s,t## are positive.

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).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##?

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.You could really try to pass all equations to their remainders modulo ##7## and look for common divisors.

- #118

fresh_42

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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

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

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##.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).

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- #119

fresh_42

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I skip the part here that is a bit confusing and unnecessary. We can always transform ##2a^6+3b^6=c^3## into ...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

$$

where we may assume w.l.o.g. ##r,s,t > 0## and ##r,s,t## are pairwise coprime integers (cp. post #118).$$

\implies 2r^3 + 3s^3=t^3

$$

Obsolete, if we first guarantee that ##s## and ##t## are coprime.With algebra, we can show that

$$

f(s,t)=\sqrt[3] \frac{t^3-3s^3}{2}

$$

If ##t,s## are both even...

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.... 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}##.

Keep it simple and only use what you have, step by step.

... which again is obsolete, since ##s## and ##t## can be assumed coprime.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##...

Summary: We are left with ##s\neq t##, both odd, and...##(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)

$$

$$

\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

$$

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.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

$$

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- #120

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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.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?

- #121

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These types of questions are great for a physics major. I'd love to look over some of the questions thanks.I got the answer as ##e^{-2}##

First, I took derivative of ##x^a e^{2a-x}## and equated it with 0, and got that maximum happens when x=a or when ##f(a)=(ae)^a## . This is ofcourse, not the maximum when a is an even number.

Then, have to find the minimum of ## g(x)=(xe)^x ## for which we use the same procedure to get ##a=e^{-2}##.

- #122

fresh_42

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In case you want to see them (at least many of them) gathered in a document, seeThese types of questions are great for a physics major. I'd love to look over some of the questions thanks.

https://www.physicsforums.com/threads/solution-manuals-for-the-math-challenges.977057/There should be quite a few useful theorems, equations, and inequalities.

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