It should be $a^x+b^y+c^z$.

As i remember this question can be solved by four numbers Theorem(I hope Fabriziofelen will write his solution)

This problem was proposed by oneplusone.

Fix a prime $p$ and look at $\nu_p$'s. Note $\nu_p(a) + \nu_p(b) + \nu_p(c) = 0$. Let $p$ be a prime for which $\nu_p(a) > 0$. WLOG $\nu_p(c) < 0$; then it follows $\nu_p(b) < 0$ too. Then we conclude $\boxed{y \nu_p(b) = z \nu_p(c)}$, so set $\nu_p(b) = -z'k$ and $\nu_p(c) = -y'k$ where $y' = \frac{y}{\gcd(y,z)}$ and $z' = \frac{z}{\gcd(y,z)}$. \[ 0 < \nu_p(a) = -\nu_p(b) -\nu_p(c) = k \cdot (y'+z'). \] Conclusion: whenever $\nu_p(a) > 0$ we have $y' + z'$ divides $\nu_p(a)$.

Here's an extremely long solution that I come up with in the APMO, and I think it's basically equivalent to the other solutions but whatever. (P.S. Could someone please fix up the LaTeX a bit sorry) Let \[a=\frac{k}{\ell},\ b=\frac{m}{n},\ c=\frac{\ell n}{km}\]where $\gcd(k,\ell)=\gcd(m,n)=1$. Then we are given that \[\frac{k^x}{\ell^x}+\frac{m^y}{n^y}+\frac{\ell^z n^z}{k^z m^z}\in \mathbb{Z}.\]Clearing the denominators, we find \[\frac{k^{x+z}m^z n^y+k^z \ell^x m^{y+z}+\ell^{x+z}n^{y+z}}{k^z\ell^xm^zn^y}\in \mathbb{Z}.\quad \quad (*)\]Now, notice that for this to be an integer we must have \[k^z\mid k^{x+z}m^z n^y+k^z \ell^x m^{y+z}+\ell^{x+z}n^{y+z}.\]Since $k^z$ divides $k^{x+z}m^z n^y$ and $k^z \ell^x m^{y+z}$, it follows that $k^z\mid \ell^{x+z}n^{y+z}$. Since $\gcd(k,\ell)=1$ it then follows that $\boxed{k^z\mid n^{y+z}}$ (1). Similarly, we can obtain the other three divisibility statements: \begin{itemize} \item $m^z\mid \ell^{x+z} \quad \quad (2)$ \item $\ell^x\mid m^zn^y \quad \quad (3)$ \item $n^y\mid k^z\ell^x \quad \quad (4)$. \end{itemize} From here, if $\gcd(\ell,n)=1$ then we are good to go. Otherwise, consider a prime $p$ that divides both $\ell$ and $n$. Suppose that \[p^\alpha\mid \mid \ell,\ p^\beta\mid \mid n.\]Then since $p\nmid k,m$, from $(3)$ we obtain \[\alpha x\leq \beta y,\]while $(4)$ implies that \[\beta y\leq \alpha x.\]From this, we can conclude that $\alpha x=\beta y$. This means that there exists positive integers $s,t$ such that $\ell=\ell'\cdot s$, $n=n'\cdot t$, $t^y=s^x$ and $\gcd(\ell',n')=1$. Indeed, if there is a common prime $p$ that divides both $\ell'$ and $n'$ at some stage, then we may factor out $p^{\frac{z}{\gcd(x,z)}}$ from $\ell'$ and $p^{\frac{x}{\gcd(x,z)}}$ from $n'$. Repeat until they do not share a common prime divisor. From now we work with $\ell'$ and $n'$ instead of $\ell$ and $n$: $(3)$ turns into $(\ell')^x\mid m^z$ and similar for $(4)$. Now, using the modified versions of (2) and (3) we can obtain $(\ell')^x\mid m^z\mid (\ell')^{x+z}$. This implies that $\ell'$ and $m$ have the same set of prime divisors. Hence, we can factor these in prime numbers and let \[\ell'=\prod^N_{j=1}p_j^{\alpha_j}\ \text{and}\ m=\prod^N_{j=1}p_j^{\beta_j}.\]Notice that if $p_j\mid s$ for some $j$, then we can obtain \[p_j\mid t\ \Rightarrow\ p_j\mid n\]which we know is bad since $\gcd(m,n)=1$. Hence, it follows that $\gcd(\ell',s)=1$ and similarly $\gcd(n',t)=1$ too. This allows us to calculate the following: \begin{itemize} \item $\nu_{p_j}(k^{x+z}m^zn^y)=\nu_{p_j}(m^z)=\boxed{z\cdot \beta_j}$ ; \item $\nu_{p_j}(k^z\ell^xm^{y+z})=\nu_{p_j}((\ell')^x\cdot m^{y+z})=\boxed{x\cdot \alpha_j+(y+z)\beta_j}$ ; \item $\nu_{p_j}(\ell^{x+z}n^{y+z})=\nu_{p_j}((\ell')^{x+z})=\boxed{(x+z)\alpha_j}$ ; \item $\nu_{p_j}(k^z\ell^xm^zn^y)=\nu_{p_j}((\ell')^x\cdot m^z)=\boxed{x\cdot \alpha_j+z\cdot \beta_j}$ . \end{itemize} Recall that if $\nu_p(u)\neq \nu_p(v)\neq \nu_p(w)$ for some integers $u,v,w$ and a prime $p$, then \[\nu_p(u+v+w)=\min\{\nu_p(u),\nu_p(v),\nu_p(w)\}.\]Noting that $x\cdot \alpha_j+(y+z)\beta_j>z\cdot \beta_j$, if we can have both the negations $z\cdot \beta_j\neq (x+z)\alpha_j$ and $(x+z)\alpha_j\neq x\cdot \alpha_j+(y+z)\beta_j$ then by the above and taking $(*)$ into consideration we can obtain \[x\cdot \alpha_j+z\cdot \beta_j\leq\min\{z\cdot \beta_j,(x+z)\alpha_j\}.\]However, this is clearly absurd, since \[\min\{z\cdot \beta_j,(x+z)\alpha_j\}\leq z\cdot \beta_j<x\cdot \alpha_j+z\cdot \beta_j.\]Here, we have used the fact that $\min\{u,v\}\leq u,v$. This implies that at least one of the equations, $z\cdot \beta_j=(x+z)\alpha_j$ and $(x+z)\alpha_j=x\cdot \alpha_j+(y+z)\beta_j$, must be true. If we have $(x+z)\alpha_j=x\cdot \alpha_j+(y+z)\beta_j>z\cdot \beta_j$ then the above contradiction still holds because \[\min\{z\cdot \beta_j,\ (x+z)\alpha_j,\ x\cdot \alpha_j+(y+z)\beta_j\}=z\cdot \beta_j<z\cdot \beta_j+x\cdot \alpha_j\]still remains true. Hence, it must follow that \begin{align*} z\cdot \beta_j&=(x+z)\alpha_j\\ \Leftrightarrow\ [p_j^{\beta_j}]^z&=[p_j^{\alpha_j}]^{x+z}\\ \Leftrightarrow\ \left[\prod^N_{k=1}p_j^{\beta_j}\right]^z&=\left[\prod^N_{k=1}p_j^{\alpha_j}\right]^{x+z}\\ \Leftrightarrow\ m^z&=(\ell')^{x+z}. \end{align*}This is enough to prove that $b$ is powerful, because $x+z>1$, implying that $m$ must be a nontrivial perfect power, and $\gcd(m,n)=1$. Similarly, we can also prove that \[k^z=(n')^{y+z},\]implying that $a$ is powerful. It only remains to prove that $c$ is powerful. In order to do this, notice that \[\frac{(\ell n)^z}{(km)^z}=\frac{(\ell n)^z}{(\ell')^{x+z}(n')^{y+z}}=\frac{s^z\cdot t^z}{(\ell')^x (n')^y},\]which shows that $\sqrt[z]{(\ell')^x (n')^y}$ is an integer, and this must be coprime to the numerator as discussed earlier. Then it follows \[s\cdot t=s\cdot s^{x/z}=[\sqrt[\frac{z}{\gcd(x,z)}]{s}]^{\frac{x+z}{\gcd(x,z)}}\]and since we can prove that $\sqrt[\frac{z}{\gcd(x,z)}]{s}$ is an integer using $t^y=s^x$, and that $x+z>\gcd(x,z)$, this expression must be a nontrivial perfect power, proving that $c$ is powerful. Having proven that $a,b,c$ are all powerful, we are done. $\square$

Who has solution using four numbers theorem

Let $a = \frac{i}{j}, b=\frac{k}{l}, c=\frac{m}{n}$ where $(i,j)=(k,l)=(m,n)=1$. Then $ikm=jln \dots (1)$ $a^x+b^y+c^z = \frac{i^xl^yn^z + j^xk^yn^z+j^xl^ym^z}{j^xl^yn^z}$ Hence $l^y \mid k^y j^x n^z \Longrightarrow l^y \mid j^xn^z \dots (2) $ Similarly $n^z \mid j^x l^y \dots (3)$ Let $p$ be a prime divisor of $i$. From equation (2) $yv_p(l)=v_p(l^y) \leq v_p(j^xn^z) = xv_p(j) +zv_p(n)$ From equation (3) $zv_p(n) = v_p(n^z) \leq v_p(j^x l^y) = xv_p(j) +lv_p(y)$ But $v_p(j)=0$ since $i,j$ are coprime. Hence $yv_p(l)=zv_p(n)$. Lastly, from $(1)$, $v_p(i) = v_p(l) +v_p(n)$. Because, if $p \mid l$, $p \mid n$ then $p \nmid k$ and $p \nmid l$. But $v_p(l) + v_p(n) = v_p(l) \bigg(1+\frac{y}{z} \bigg) = v_p(l) \bigg( \frac{y+z}{y} \bigg) = \frac{v_p(l)}{\frac{y}{(y,z)}} \bigg(\frac{y+z}{(y,z)} \bigg)$ All prime exponents of $i$ are divisible by $\alpha = \frac{y+z}{(y,z)} \geq 2$ so $i$ is a perfect $\alpha$ power. Similarly $k,m$ are perfect powers. We are done.

v_Enhance wrote: Fix a prime $p$ and look at $\nu_p$'s. Note $\nu_p(a) + \nu_p(b) + \nu_p(c) = 0$. Let $p$ be a prime for which $\nu_p(a) > 0$. WLOG $\nu_p(c) < 0$; then it follows $\nu_p(b) < 0$ too. Then we conclude $\boxed{y \nu_p(b) = z \nu_p(c)}$, so set $\nu_p(b) = -z'k$ and $\nu_p(c) = -y'k$ where $y' = \frac{y}{\gcd(y,z)}$ and $z' = \frac{z}{\gcd(y,z)}$. \[ 0 < \nu_p(a) = -\nu_p(b) -\nu_p(c) = k \cdot (y'+z'). \] Conclusion: whenever $\nu_p(a) > 0$ we have $y' + z'$ divides $\nu_p(a)$. Why does $v_p(b) < 0$ if $v_p(a) > 0$ and $v_p(c) < 0$

The situation $\nu_p(a) > 0$ and $\nu_p(b) \ge 0$ and $\nu_p(c) < 0$ is impossible because it gives $\nu_p(a^x+b^y+c^z) = \nu_p(c^z) < 0$.

Say we had a prime $p$ such that $\nu_p(a)=0$, so $\nu_p(b)=-\nu_p(c)$. Then, $\nu_p(a^x+b^y+c^z)$ would be negative unless $\nu_p(b)=\nu_p(c)=0$. Thus, we can't have any of $(\nu_p(a),\nu_p(b),\nu_p(c))$ be $0$ unless they are all $0$. Suppose $\nu_p(a)<0$ and $\nu_p(b),\nu_p(c)>0$. Then, we have a problem as $\nu_p(a^x+b^y+c^z)=x\nu_p(a)<0$. Thus, we must have exactly two negatives and one positive in $(\nu_p(a),\nu_p(b),\nu_p(c))$. Suppose $p$ is some prime such that $\nu_p(a)>0$, so from above, we see that we must have $\nu_p(b),\nu_p(c)<0$. If $y\nu_p(b)\ne z\nu_p(c)$, then \[\nu_p(a^x+b^y+c^z)=\min\{y\nu_p(b),z\nu_p(c)\}<0,\]so we must in fact have $y\nu_p(b)=z\nu_p(c)$. Thus, $(\nu_p(b),\nu_p(c))=-k(z',y')$ where $k\in\mathbb{N}$ and $(y',z')=\frac{(y,z)}{\gcd(y,z)}$, so $\nu_p(a)=k(y'+z')$. Thus, if $\nu_p(a)>0$, then $y'+z'\mid a$, so the numerator of $a$ is a perfect power, as desired.

I claim that $c$ can be written in the form $\frac{m^{\frac{x+y}{\gcd(x,y)}}}{n}$, with $\gcd(m,n)=1$, which implies $c$ is powerful, as $\frac{x+y}{\gcd(x,y)}\ge 2$. Consider a prime $p|c$, and $v_p(a),v_p(b)$. As $v_p(a)+v_p(b)+v_p(c)=0$, at least one of them must be negative. However, if only one value, WLOG $v_p(a)$ is negative, then $v_p(a^x+b^y+c^z)=xv_p(a)<0$. Hence, we must have $v_p(a),v_p(b)<0$. Furthermore, $v_p(a^x+b^y+c^z)\ge \min(xv_p(a),yv_p(b))$, with the inequality strict iff $xv_p(a)=yv_p(b)$. As $a^x+b^y+c^z$ is an integer, we must have the latter equality, so we are able to write $v_p(a),v_p(b)$ as $-\frac{ky}{\gcd(x,y)},-\frac{kx}{\gcd(x,y)}$ respectively, for some $k$. So, $v_p(c)=k\left(\frac{x+y}{\gcd(x,y)}\right)$. Repeating this argument for all prime factors of $c$, we see that all $v_p$s are divisible by $\frac{x+y}{\gcd(x,y)}$, thus implying the claim. As this argument can be repeated for all $3$ numbers, we have $a,b,c$ are all powerful, as desired.

Does there exist a set of numbers satisfying the conditions besides $1,1,1$?

Let $a=\frac{p_1}{q_1},b=\frac{p_2}{q_2},c=\frac{p_3}{q_3}$, such that $(p_i,q_i)=1$, then we have $$p_1^xq_2^yq_3^z+q_1^xp_2^yq_3^z+q_1^xq_2^yp_3^z=q_1^xq_2^yq_3^z\hspace{20pt}(1)$$and $$p_1p_2p_3=q_1q_2q_3\hspace{20pt}(2)$$Claim 1. $(p_1,p_2)=(p_2,p_3)=(p_3,p_1)=1$ Proof. Suppose $r|p_1,p_2$, thenby $(2)$ we have $r|q_3$ and $r\nmid p_3,q_1,q_2$. Therefore, in $(1)$, the third term is not divisble by $r$ while all other does, contradiction. $\blacksquare$ Claim 2. Let $(y,z)=d$, $ed=y$ and $df=z$. If $p$ is a prime such that $p|p_1$ then $e+f|v_p(p_1)$ Proof. From Claim 1, $v_p(p_2)=v_p(p_3)=v_p(q_1)=0$. Therefore, letting $v_p(p_1)=a,v_p(q_2)=b,v_p(q_3)=c$, among the four terms in (1), $$v_p(p_1^xq_2^yq_3^z)=xa+yb+zc, v_p(q_1^xp_2^yq_3^z)=zc, v_p(q_1^xq_2^yp_3^z)=yb,v_p(q_1^xq^2yq^3z)=zc+yb$$Hence $zc=yb$, $eb=fc$,hence $e|c$ and $b|f$ $$a=b+c=\frac{fc}{e}+c=\frac{c(f+e)}{e}$$as desired. $\blacksquare$ Therefore $a=\frac{p_1}{q_1}=\frac{p^{y+z}}{q_1}$ for some $p$ so $a$ is powerful, similarly $b,c$ are powerful as well.

Can anybody check this for me please.

Throughout this solution, when I refer to the numerator and denominator of a rational fraction, assume that they are reduced. I will first show that $a$ is powerful. Let $p$ be a fixed prime dividing the numerator of $a$, so $\nu_p(a)>0$. If no such prime exists, then $a$ is great as its numerator is necessarily $1$, which is a perfect power. Since $\nu_p(a)+\nu_p(b)+\nu_p(c)=0$, it follows that at least one of $\nu_p(b),\nu_p(c)$ must be negative. WLOG let $\nu_p(b)<0$. Then $\nu_p(c)<0$ as well, since if $\nu_p(c) \geq 0$ then $\nu_p(b^y)$ is negative, while $\nu_p(a^x+c^z)$ is positive. If $y\nu_p(b)\neq z\nu_p(c)$, then we have $\nu_p(b^y+c^z)=\min\{y\nu_p(b),z\nu_p(c)\}<0$, so $a^x+b^y+c^z$ cannot be an integer. Hence we obtain that $\boxed{y\nu_p(b)=z\nu_p(c)}$, so $\frac{\nu_p(b)}{\nu_p(c)}=\frac{z}{y}$. Since $p$ is arbitrary, this means that $\frac{\nu_p(b)}{\nu_p(c)}$ is equal to some positive rational constant $k$ over all primes $p$ dividing the numerator of $a$. Hence: $$\frac{\nu_p(b)}{\nu_p(c)}=k \implies \nu_p(b)=k\nu_p(c) \implies \nu_p(a)=-(\nu_p(b)+\nu_p(c))=-(k+1)\nu_p(c).$$Let $n$ be the least positive integer such that $(k+1)n$ is an integer. It is clear that $\nu_p(c)$ must be a multiple of $n$ for all $p$, hence $(k+1)n \mid \nu_p(a)$ for all $p$. Since $k+1>1$, it follows that $(k+1)n$ is at least two, so the numerator of $a$ can be written as some positive integer raised to the power of $(k+1)n>1$. Hence $a$ is powerful. By symmetry, this also implies that $b,c$ are powerful. $\blacksquare$

dame dame

Let $a=\dfrac{n_a}{d_a}$, $b=\dfrac{n_b}{d_b}$ and $c=\dfrac{n_c}{d_c}$ expressed as reduced in lowest terms. From $a^x + b^y + c^z$ being integer, we obtain that \begin{align*} d_c^z\mid d_a^xd_b^y\\ d_b^y\mid d_a^xd_c^z\\ d_a^x\mid d_b^yd_c^z.\\ \end{align*}Now we look $v_p$'s. Let $p\mid n_a$, then $v_p(n_a)>0$ and $v_p(d_a)=0$, also if $v_p(d_b)=0$, then $v_p(d_c)=0$ by divisibility condition above, but this contradicts $abc=1$ condition. Therefore both $v_p(d_b)>0$ and $v_p(d_c)>0$. Furthermore from the divisibilities we get that $zv_p(d_c)=yv_p(d_b)$, so we rewrite this as $v_p(d_c)=\frac{z}{\gcd(y,z)}k$ and $v_p(d_b)=\frac{y}{\gcd(y,z)}k$, thus $v_p(n_a)=\left(\frac{y}{\gcd(y,z)}+\frac{z}{\gcd(y,z)}\right)k$. This means that for every prime we obtain the exponent $\frac{y}{\gcd(y,z)}+\frac{z}{\gcd(y,z)}\geq 2$ for $n_a$, we do the same for $n_b$, $n_c$, we are done.

For storage. By symmetry, it only suffices to show that one of $a$, $b$, or $c$ is powerful. For any prime $p$, we have $\nu_p(a)+\nu_p(b)+\nu_p(c) = 0$. Let $p$ be an arbitrary prime such that $\nu_p(a) > 0$. Clearly, at least one of $\nu_p(b)$ or $\nu_p(c)$ is negative. In the case that they are of opposite signs, we have $\nu_p(a^x + b^y + c^z) = \min\{x\nu_p(a),y\nu_p(b),z\nu_p(c)\} < 0$ which implies that the expression is not an integer. Hence we have that they are both negative. Now, given that $v_p(a^x + b^y + c^z)$ is non-negative, we have that $x$ and $y$ satisfies $y\nu_p(b) = z\nu_p(c)$. If we let $y' = \frac{y}{\gcd(y,z)}$ and $z' = \frac{z}{\gcd(y,z)}$, then we have that $y'\nu_p(b) = z'\nu_p(c)$. Now, because $\gcd(y',z')=1$ we now obtain that $\nu_p(b) = -kz'$ and $\nu_p(c) = -ky'$ for some positive integer $k$. In particular, $\nu_p(a) = k(y'+z')$, which means that $y' + z' \mid \nu_p(a)$, hence we're done as this implies that the numerator of $a$ is a perfect $y' + z'$th power and thus $a$ is powerful.

It suffices to prove $a$ is powerful; $b$ and $c$ follow similarly. Consider a prime $p$ such that $\nu_p(a)\ge 0.$ We have $\nu_p(b)=\nu_p(c)=0$ or $\nu_p(b)\ge 0$ and $\nu_p(c)<0$ (or vice versa) or $\nu_p(b)<0$ and $\nu_p(c)<0.$ In the first case, $a=b=c=1,$ which are all powerful. In the second case, $$\nu_p(a^x+b^y+c^z)=\min\{\nu_p(a^x+b^y),\nu_p(c^z)\}=z\nu_p(c)<0$$which is absurd. In the third case, $$\nu_p(a^x+b^y+c^z)=\min\{\nu_p(a^x),\nu(b^y+c^z)\}.$$If $\nu_p(b^y)\neq\nu_p(c^z),$ $$\nu_p(b^y+c^z)=\min\{\nu_p(b^y),\nu_p(c^z)\}<0,$$so $\nu_p(a^x+b^y+c^z)<0.$ Hence, $y\nu_p(b)=z\nu_p(c)$ and we can let $\nu_p(b)=\frac{z}{\gcd(y,z)}k$ and $\nu_p(c)=\frac{y}{\gcd(y,z)}k.$ Then, $$\nu_p(a)=-\nu_p(b)-\nu_p(c)=k\left(\frac{y+z}{\gcd(y,z)}\right)$$so $\frac{y+z}{\gcd(y,z)}\mid\nu_p(a).$ This is true for all $p$ such that $\nu_p(a)>0$ so $a$ is powerful. $\square$

gghx wrote: Does there exist a set of numbers satisfying the conditions besides $1,1,1$? $(a,b,c,x,y,z)=\left(\frac{3^2}{2},\frac{2^2}{3},\frac{1}{6},1,1,1\right)$

Just analyze $a, b, c = \frac{p}{q}, \frac{q}{r}, \frac{r}{p}$ with $\gcd(p, q, r) = 1$ to get that $\frac{r}{\gcd(r, p)} = n^{\frac{x+y}{\gcd(x, y)}}$ and others are from $\ell, m$ where $n, \ell, m$ are pairwise coprime. This finishes. Similar to the problem of determining $\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \in \mathbb{Z}$.

I assume the problem proposers must have a solution where $k = 2$ does not suffice? I wonder what it is.

I suppose a,b,c=1,1,1 technically doesn't work; 3 is not a perfect power but whatever. Let a,b,c be distinct, and take $(d_1,d_2,...),(e_1,e_2,...),(f_1,f_2,...):v_{p_1}(a)=d_1$, etc. For convenience of typing, d,e,f represent arbitrary terms in their respective sequences $d_i,e_i,f_i$. Since d+e+f=0, we know two of them must be negative (otherwise either they're all equal which we can disregard since they're all 0, or two positive one negative implies $a^x+b^y+c^z$ is not an integer); in particular, WLOG d,e<0, we must have $$dx=ey\implies\exists s=\frac x{\gcd(x,y)},t=\frac y{\gcd(x,y)}:d=-sk,e=-tk\implies f=-d-e=k(s+t),$$which is an integer multiple of k. edit: oops apparently the term for it is powerful but in otis theres no term called that lol @below oopsies

huashiliao2020 wrote: I suppose a,b,c=1,1,1 technically doesn't work; 3 is not a perfect power but whatever. $a,b,c$ are powerful, not $a+b+c$..?

Wow. I solved it while I was watching anime. Let $p$ be a prime. Then $\nu_{p}(a) + \nu_{p}(b) + \nu_{p}(c) = 0$. Assume $\nu_{p}(a), \nu_{p}(b), \nu_{p}(c)$ are not all 0. Assume $\nu_{p}(b), \nu_{p}(c) \geq 0$ and $\nu_{p}(a) < 0$. Then $\nu_{p}(b^y + c^z) \geq 0 > \nu_{p}(a^x)$, so $\nu_{p}(a^x + b^y + c^z) < 0$, a contradiction. So WLOG we can assume $\nu_{p}(b), \nu_{p}(c) < 0$ and $\nu_{p}(a) > 0$. Then we have $\nu_{p}(b^y) = \nu_{p}(c^z)$, otherwise $0 \geq \nu_{p}(a^x + b^y + c^z) = \min(\nu_{p}(a^x), \nu_{p}(b^y), \nu_{p}(c^z)) > 0$, a contradiction. Thus $\nu_{p}(a)$ is divisible by $\frac{x + y}{\gcd (x, y)}$. This completes proof.

$\color{magenta}\boxed{\textbf{SOLUTION P4}}$ $\color{red}\textbf{Note :}$ $$\nu_{p}(a)+\nu_{p}(b)+\nu_{p}(c)=0$$And $$\nu_{p}(a^{x})+\nu_{p}(b^{y})+\nu_{p}(c^{z} \ge 0$$ $\color{green} \textbf{Claim 1 :}$ There exist $p$ such that $\nu_{p}(a) > 0$ $\textbf{Proof :}$ Assume not, So for all $p, \nu_{p}(a)\le 0 \implies a=1/p^{m}a$ which is powerful $\square$ $\color{green} \textbf{Claim 2 :}$ $\nu_{p}(b) < 0, \nu_{p}(c) < 0$ $\textbf{Proof :}$ If not, We have two cases $\textbf{1.}$ $\nu_{p}(b)=\nu_{p}(c)=0 \implies a=b=c=1$ which are all powerful $\textbf{2.}$ $$\nu_{p}(b)<0, \nu_{p}(c)>0 \implies \nu_p(a^x+b^y+c^z)=\min\{\nu_{p}(a^x+b^y),\nu_{p}(c^z)\}=z\nu_{p}(c)<0$$which is not possible Hence done $\square$ $\color{green} \textbf{Claim 3 :}$ $\nu_{p}(b^y)=\nu_{p}(c^z)$ $\textbf{Proof :}$ $$\nu_{p}(a^x+b^y+c^z)=\min\{\nu_{p}(a^x),\nu_{p}(b^y+c^z)\}$$$$\nu_{p}(b^y)\neq\nu_p(c^z) \implies \nu_{p}(b^y+c^z)=\min\{\nu_{p}(b^y),\nu_{p}(c^z)\}<0$$$$\implies \nu_{p}(a^x+b^y+c^z)<0$$Which is not possible $\square$ Now, Let $\gcd(y,z)=d$ and $\nu_{p}(b)=-z'k, \nu_{p}(c)=-y'k$ where $y'=\frac{y}{d}, z'=\frac{z}{d}$ So, $$ 0 < \nu_p(a) = -\nu_{p}(b) -\nu_{p}(c) = k(y'+z')$$$$\implies y'+z'|\nu_{p}(a), y'+z'\ge 2$$$$\implies a=\frac{p^{y'+z'}}{q}\blacksquare$$

Let $p$ be a prime such that $\nu_p(a)>0$. And so we have that \[\nu_p(a)+\nu_p(b)+\nu_p(c)=0 \text{ and } \nu_p(a^x+b^y+c^z) \geq 0\]WLOG say $\nu_p(b) \geq 0$, then $\nu_p(c)<0$ and so we have \[0 \leq \nu_p(a^x+b^y+c^z)=z\nu_p(c)<0\]which is a contradiction. And hence $\nu_p(b)$, $\nu_p(c)<0$. WLOG say $y\nu_p(b)<z\nu_p(c)$ and so again we get \[0 \leq \nu_p(a^x+b^y+c^z)=y\nu_p(b)<0\]again contradiction. And so we have \[y\nu_p(b)=z\nu_p(c) \text{ and } \nu_p(a)=-\nu_p(b)-\nu_p(c) \implies \left. \frac{y+z}{\gcd(y,z)} \right| \nu_p(a) \]And so we are done as $\frac{y+z}{\gcd(y,z)}>1$.

Let $p$ be any prime such that $v_p(a)>0$. Then $v_p(b)+v_p(c)=-v_p(a)$. If $v_p(b)>0$, then $v_p(c)<0$, which is impossible since $a^x+b^y+c^z$ is an integer. So we have $v_p(b), v_p(c)<0$. We also have $yv_p(b)=zv_p(c) \implies v_p(b)=-\frac{nz}{gcd(y,z)},v_p(c)=-\frac{ny}{gcd(y,z)}, v_p(a)=\frac{n(y+z)}{gcd(y,z)}$ for some positive integer $n$. Write $x_y=\frac{x}{gcd(x,y)}$, and let $x_z,y_x,y_z,z_x,z_y$ similarly. It follows that $a=\frac{s^{y_z+z_y}}{t^{z_x}u^{y_x}},b=\frac{t^{x_z+z_x}}{s^{z_y}u^{x_y}},c=\frac{u^{x_y+y_x}}{s^{y_z}t^{x_z}}$ for some pairwise coprime positive integers $s,t,u$. Now $y_z+z_y=\frac{y+z}{gcd(y,z)}$ which is clearly larger than $1$, so $a$ is powerful. The rest follows similarly. $\square$