A remark: The original statement asks to find all positive integers $m$ so that there are real numbers $a_1,\ldots, a_m$ satisfying the equations.

Consider $b_i=a_i+0.5$, we have $$\sum b_i=\frac{3m}{2},\quad\sum b_i^2=\frac{49m}{4},\quad\sum b_i^3=\frac{147m}{8},\quad\sum b_i^4=\frac{2401m}{16}$$by simple calculation. Note that $$m\left(\sum b_i^4\right)=\left(\sum b_i^2\right)^2. $$By Cauchy’s inequality, $b_i^2$ is constant in $i$, which is $49/4$. Let $p$ be the number of $i$ such that $b_i=7/2$, and $q=m-p$ be the number of $i$ such that $b_i=-7/2$. Then $$7(p-q)=3m.$$Hence $m$ is a multiple of $7$.

Untro368 wrote: Note that $$\left(\sum b_i^4\right)\left(\sum b_i^2\right)=\left(\sum b_i^3\right)^2. $$ I think it should be $m\left(\sum b_i^4\right)=\left(\sum b_i^2\right)^2$, from which we can get that $b_i^2$ is constant in $i$.

Scrutiny wrote: Untro368 wrote: Note that $$\left(\sum b_i^4\right)\left(\sum b_i^2\right)=\left(\sum b_i^3\right)^2. $$ I think it should be $m\left(\sum b_i^4\right)=\left(\sum b_i^2\right)^2$, from which we can get that $b_i^2$ is constant in $i$. Yes, you are right.

Untro368 wrote: Consider $b_i=a_i+0.5$, we have $$\sum b_i=\frac{3m}{2},\quad\sum b_i^2=\frac{49m}{4},\quad\sum b_i^3=\frac{147m}{8},\quad\sum b_i^4=\frac{2401m}{16}$$by simple calculation. Note that $$m\left(\sum b_i^4\right)=\left(\sum b_i^2\right)^2. $$By Cauchy’s inequality, $b_i^2$ is constant in $i$, which is $49/4$. Let $p$ be the number of $i$ such that $b_i=7/2$, and $q=m-p$ be the number of $i$ such that $b_i=-7/2$. Then $$7(p-q)=3m.$$Hence $m$ is a multiple of $7$. Once knowing that the $a_i$'s are supposed to be supported on $3$ and $-4$, we could actually just compute and show that $\sum_{i=1}^{m}(a_i-3)^2(a_i+4)^2=0$ and be done.

funny problem Let $b_i = a_i + \frac{1}{2}$ notice $\sum b_i^2 = \frac{49}{4}, \sum b_i^4 = \frac{2401}{16}$ so by cauchy $b_i = \pm \frac{7}{2}$, so the the $a_i$ consists of $-4$s and $3$s, Let $p$ and $q$ be the number of $-4,3$ respectively. so $3m = 3p+3q = 7(p-q) $ so done.

Remark

@below, The no. were weird so initially I used $b_i = a_i+k$ then I thought $\frac{1}{2}$ could be good (because of the squares).

Just out of curiosity, how would you come up with the substitution $b_i=a_i-\frac{1}{2}$ without already knowing the equality case?

For this type of question, i guess, it is usually some inequality hidden in the equality or the sum of squares. So the equation must be solvable by hand. Checking $m=7$ may help

KI_HG wrote: For this type of question, i guess, it is usually some inequality hidden in the equality or the sum of squares. So the equation must be solvable by hand. Checking $m=7$ may help Yes, I mean I solved the problem myself by looking for integer solutions when $m=7$ and then you discover $3,-4$ and the rest is easy as described in #5. Again, my question was whether there is a way to come up with the argument without first guessing the solutions for $m=7$.

OK so here is an "ex post" rationalization of how one could come up with such a solution, without first considering the solutions for the case $m=7$: It is natural to look at inequalities, and so comparing sums of squares and fourth powers, we would get a contradiction from Cauchy if $11^2>131$. Unfortunately, this is not quite the case, but it seems to fail only barely. So what could we try? Well, if we substitute $b_i=a_i+c$ for some constant $c$ to be specified later, we still get a system of equations, computing $\sum b_i, \sum b_i^2, \sum b_i^3, \sum b_i^4$. So we can try to use Cauchy on the $b_i$. Precisely, we get $\frac{1}{m}\sum b_i^2=11+2c+c^2$ and $\frac{1}{m} \sum b_i^4=131+4c+66c^2+4c^3+c^4$. So we would win (i.e. get a contradiction from Cauchy) if we could find $c$ such that $(c^2+2c+11)^2 >c^4+4c^3+66c^2+4c+131$. Expanding this and cancelling, we end up with the equivalent $0>40c^2-40c+10=10(2c-1)^2$. So it will not be possible, but almost: If we choose $c=\frac{1}{2}$, we get the equality case and hence all the $b_i^2$ must be the same in this case, i.e. they must all be $11+2c+c^2=12.25$ i.e. all the $b_i$ are $\pm 3.5$ i.e. all the $a_i$ are $-4$ or $3$, and the rest is easy.

We uploaded our solution https://calimath.org/pdf/TaiwanMO2023-5.pdf on youtube https://youtu.be/6dsV22rOXU8.