Here's a solution. First, add and subtract the equations to get $(x^3+y+1)^2-1+z^9=147^{157}-157^{147}$ $(x^3-y)(x^3+y)-z^9=147^{157}-157^{147}$ Now, assume that $x,y,z$ are a solution and work mod 19. $147^{157}\equiv2$ and $157^{147}\equiv11,$ so $(x^3+y+1)^2+z^9\equiv14$, Since 9th powers are congruent to $\pm1$, either $(x^3+y+1)^2\equiv13$ or $(x^3+y+1)^2\equiv15$. Neither of these values is a square mod 19, so there is no solution and we are done. Edit- I'm having trouble calculating. My previous argument used the wrong value for $147^{157}$. Since it worked, I didn't realize it was wrong at the time.

One more correction: 0 is also a 9th power, 14 is also not a square. I don't know what it is about this problem, but I just can't calculate straight.

The other viable solution which I used when I took the contest was considering things modulo 13. Basically, the only possible residues mod 13 for cubes (and ninth powers) are 0, 1, 5, -1, -5, so by plugging these in as possible values for $x^3$ in the first system are those. Plugging them in gives corresponding values for $y$, but plugging them in to the second equation produces no viable solutions for $z^9$.

jmerry wrote: One more correction: 0 is also a 9th power, 14 is also not a square. I don't know what it is about this problem, but I just can't calculate straight. What was the rationale between choosing 13 and 19 as mods? Did you just keep trying ones and these had nice residues, or was there something that led you to use them? My first intuition was mod 3 and 7, then CRT into mod 21 and reach a contradiction (which worked, but took about 2.5 hours to get; luckily #1 was trivialized by massive Graph theory).

In general, when you want to minimize the number of residues of a power $k$ in a modular congruence $m$, then you want to choose an $m$ such that $k|\phi(m)$. In this case, $k=9$, so we choose $19$ because $\phi(19)=18$, and $9|18$. For $13$, it's less obvious.

This problem is nice we had solved it with our teacher... Davron Latipov

Elemennop wrote: In general, when you want to minimize the number of residues of a power $k$ in a modular congruence $m$, then you want to choose an $m$ such that $k|\phi(m)$. In this case, $k=9$, so we choose $19$ because $\phi(19)=18$, and $9|18$. For $13$, it's less obvious. where can i get a file or a link to study this method ? Sincerely Davron

Hmm.. I was doing this problem as practice and ended up with a really long winded solution, which I believe works but would like to confirm this if anyone would be so kind. Consider equation 1, factoring we have: $ (x^3+y)(x^3+1) = 147^{157}$ Equation 2 factors to: $ (x^3+y)(y+1) = 157^{147}-z^9$ Since both LHS's have a common term, we now examine it. In particular, we know $ x^3+y|147^{157}$ And so we have 3 cases: Case 1: $ x^3+1=\pm 1$ Then $ \pm(x^3+1)=147^{157}$ So either $ 147^{157}+1,147^{157}-1$ is a perfect cube. In addition, 147 is odd, so both of those terms are even, so they must be divisible by 8. However, it is easy to check that neither one is ($ 147^{157} \equiv 147 \equiv 3 \pmod 8$). Case 2: $ 7|x^3+y$ Plugging this into the second equation, we see that $ 7|157^{147}-z^9 \Rightarrow z^9 \equiv 157^{147} \Rightarrow z^3 \equiv 157^{49} \equiv 3^{49} \equiv 3 \pmod 7$ However, any perfect cube modulo 7 must be congruent to one of $ -1,0,1$ by Fermat, so we have a contradiction. Case 3: $ 3|x^3+y$ This is the most painful case... but we will proceed with algebraic manipulations on equation 1: ${ \pm 3^k(x^3+1)=147^{157} x^3+1=\pm \frac{146^{157}}{3^k}}$ So there must exist a perfect cube of the form: $ 3^{157-k}7^{2*157} \pm 1$ This must be even, so must be divisible by 8, and we have $ 3^{157-k}7^{2*157} \equiv 3^{\{0,1\}}$ Thus, if we are to have a perfect cube, 157-k must be even, and the "$ \pm$" must denote a minus one. We now have: $ 3^{2r}7^{2*157} - 1=a^3$ for some a. This is a contradiction to Mihailescu's Theorem so we're done. Did I screw up stupidly anywhere?

Hi I was just wondering; how do you think of using $\pmod{13}$? Also for the CRT solution that @JSteinhardt mentioned, would you just find all possibilities for $x,y,z$ $\pmod{3}$ and $\pmod{7}$ and then CRT them and show that the resulting $\pmod{21}$ congruences don't work?

i used mod 9 and it seems to have worked fine...

Add the two equations and work modular $19$. We get $(x^3+y+1)^2+z^9 \equiv 14 \pmod{19}$, and $z^9 \equiv 0, \pm 1 \pmod{19}$. However, we can check the quadratic residues $\left(\frac{13}{19}\right), \left(\frac{14}{19}\right), \left(\frac{15}{19}\right)$ so we are done.

Valentin Vornicu wrote: Prove that the system \begin{align*} x^6+x^3+x^3y+y & = 147^{157} \\ x^3+x^3y+y^2+y+z^9 & = 157^{147} \end{align*}has no solutions in integers $x$, $y$, and $z$. proving that $x^3+y=3^k$ will easily solve the question since then $x^3+1 \geq 7^{314}*3$ which is clearly bigger than $3^{147}$ so it will give us a contradiction and well proving that is just by checking modulo 7 in the second equation

How does one know they should work $mod$ $19$?

Another a lot lenghtier approach: Factoring the first equation gives you $(x^3 + y)(x^3 + 1)=147^{157}=3^{157}\cdot 7^{314} $ . Now you get that there exists nonnegative integers $a,b$ for which $(x^3+1)=\pm 3^{a}\cdot 7^{b}$, which when solving the diophantine equation gives you some little solutions for $x$($x=0,2,-2,-4$) and also the corresponding values of $y$, which are not very hard to check off using the second equation and small modulos(like 7 and 9).

SenorIncongito wrote: How does one know they should work $mod$ $19$? Well, if $ ab=p-1 $ where $p$ is an odd prime, then $a$-th powers give exactly $b$ non-zero remainders mod $p$. In particular, $9*2=19-1$, so ninth powers give only a few remainders (0,1,-1) and $z$ appears only in $z^9$, so it is reasonable to take mod 19.

yuck We actually compute the choices of $x$ with $x^3+1\mid 147^{157}$, which must occur if $147^{157}=x^6+x^3+x^3y+y = (x^3+y)(x^3+1)$. Remark that $x>1$ because $x=1$ fails. Observe that the only prime factors of $x^3+1$ are $3$ and $7$. Observe that $9\nmid x^2-x+1$ because $3\nmid x^2-x+1$ if $x\equiv 0,1\pmod 3$ and $9$ does not divide any of $2^2-2+1=3,5^2-5+1=21,8^2-8+1=57$. Remark that $3\mid x+1$ if and only if $3\mid x^2-x+1$. Moreover, \[\gcd(x+1,x^2-x+1) = \gcd(x+1,(-1)^2-(-1)+1) = \gcd(x+1,3),\]so if $3$ did not divide $x+1$ it would have to be $1$, as $x^2-x+1$ would be at least $3^2-3+1=7$ and therefore divisible by $7$ because it is not divisible by $3$. This is absurd, so $3\mid x+1$. By the gcd condition, we can write $x+1=3^p$ and $x^2-x+1=3\cdot 7^q$ for some integers $p\ge 1$ and $q\ge 0$. Write \[3\cdot 7^q = (3^p-1)^2 - (3^p-1)+1 = 3^{2p}-3^{p+1}+3.\]This implies $7^q-1 = 3^{2p-1}- 3^p = (3^{p-1}-1)3^p$. Assume for now that $q>0$. It is clear that $2p-1> q$ because otherwise $7^q = 3^{2p-1}-3^p+1 \le 3^{2p-1} \le 3^q$, absurd. But we also have $p = \nu_3(q)+1$ by LTE. Thus $2\log_e q + 1\ge 2\log_3 q+1\ge 2\nu_3(q)+1> q$. Remark that $2\log_e q+1-q$ has derivative $\frac 2q-1$, so it is decreasing if $q>2$. Moreover, the inequality does not hold at $q=5$ because we would have to have $2\log_e 5 \ge 4$ otherwise, which would be true if and only if $e^2 \le 5$. But in fact $e\ge 2.7$, so indeed the result does not hold for $q=5$. For $q=1,2,3,4$ manually check, noting that at $q=1,2,3,4$ we get $2\nu_3(q)<q$. Thus $q=0$, so $3^{p-1}-1=0$, meaning $p=1$. This yields the one solution of $x=2$. However, we did not consider the situation in which $x$ is negative or zero (in which case $x=0$ works). In that case, let $a=-x$ and observe that for essentially the same reasons as before, $a^3-1=(a-1)(a^2+a+1)$ must either have $a=2$ (which is a solution) or $a=3^p$ and $a^2+a+1 = 3\cdot 7^q$. By the same machinations as before, we get $(3^{p-1}+1)3^p = 7^q-1$. Again, we have $p = \nu_3(q)+1$. This time we remark that either $p=1$ (which yields no solution) or $7^q = 3^{2p-1}+3^p+1< 3^{2p-1}+3^{p+1} < 3^{2p}$, meaning that $q< 2p = 2\nu_3(q)+2$. That is, $q\le 2\nu_3(q)+1\le 2\log_3 q+1 \le 2\log_e q+1$. Again, $q<5$ by the argument from before, so $q\in \{1,2,3,4\}$ because $q=0$ would be absurd. Again, note that $q$ cannot be in $\{2,4\}$ because then $q\le 1=2\nu_3(q)+1$, absurd. If $q=1$ we get $p=1$ so $a=4$ and we do not have $7^3 = 3^3 + 3^2 + 1$. Thus the possible values of $x$ are $\{2,0,-2,-4\}$. Write $147^{157} = (x^3+1)(x^3+y),157^{147}=(x^3+y)(y+1)+z^9$. It is clear that $27\mid x^3+y$, so \[z^9\equiv 157^{147}\equiv 22^{147}\equiv 22^3\equiv (-5)^3 \equiv -125\equiv 10\pmod{27}.\]Since $10$ is not a $9$-th power modulo $27$, we are done.

Wait, isn't this just $\bmod{4}$ ? It's not hard to prove that $x,y,z$ are all odd. (by subtracting the first equation by the second) Hence, by Euler; \begin{align*} x+xy+y & \equiv 2 \\ x+xy+y+z & \equiv 0 \end{align*}So $z\equiv 2$ and $x+y+xy\equiv 2\implies (x+1)(y+1)\equiv 3$. This means the pair $(x+1,y+1)\equiv (1,3),(3,1)\implies x,y$ are even integers. But $x^6+x^3+x^3y+y = 147^{157}$ is not, which is a contradiction.

Add the two equations. $x^6+2x^3+2x^3y+y^2+2y+z^9=147^{157}+157^{147}$, so \[(x^3+y)(x^3+y+2)+z^9=147^{157}+157^{147}\] We work in $\pmod{19}$. First, we have $147^{157}\equiv 14^{13}\equiv 2\pmod {19}$ and $157^{147}\equiv 5^3\equiv 11\pmod{19}$. So $(x^3+y)(x^3+y+2)+z^9\equiv 13\pmod{19}$. Clearly $z^9\equiv \{-1,0,1\}\pmod{19}$. Case 1: $z^9\equiv 1\pmod{19}$. Then $(x^3+y)(x^3+y+2)\equiv 12\pmod{19}\implies (x^3+y+1)^2\equiv 13\pmod{19}$. This is a contradiction because $13$ is not a QR mod $19$. Case 2: $z^9\equiv 0\pmod{19}$. Then $(x^3+y+1)^2\equiv 14\pmod{19}$. $14$ is not QR mod $19$. Case 3: $z^9\equiv -1\pmod{19}$. Then $(x^3+y+1)^2\equiv 15\pmod{19}$. $15$ is not QR mod $19$. We have exhausted all cases, so there are no solutions.

We can show in any one preferable modular and do casework to show that there is no solution for such a system. Nonetheless I did in $\pmod {13}.$

Upon factoring, our equations become $$(x^3+1)(x^3+y)=147^{157}$$$$(y+1)(x^3+y)+z^9=157^{147}$$We work in modulo $19$. Note that $147^{157}\equiv 2$ (mod $19$) and $157^{147}\equiv 11$ (mod $19$). The cubic residues mod $19$ are $0,1,7,8,11,12,18$, so the corresponding residues of $x^3+1$ are $1,2,8,9,12,13$, since $147^{157}$ isn’t divisible by $19$. Then we must have $x^3+y$ have corresponding possible residues of $2,1,5,15,16,6$, yielding corresponding residues of $y+1$ as $3,1,18,8,6,14$. Then, in our second equation, the possible residues of $(y+1)(x^3+y)$ mod $19$ are $1,6,8,14$, so we must have $z^3$ equivalent to $3,5,10,16$ mod $19$, but since none of these are nonic residues, there are no integer solutions to our system, as desired.

Pretty weird problem. Add up both equations to get $(x^3+y+1)^2 + z^9 = 147^{157} + 157^{147} +1$. We can take both sides mod $19$. We know that $z^9 \equiv -1, 0, 1 \pmod{19}$ since $z^{18} \equiv 0,1 \pmod{19}$. The right side can be found pretty easily mod 19: $$147^{157}+157^{147} + 1 \equiv (-5)^{13} + 5^{3} +1 \equiv 2 + 11 + 1\equiv 14 \pmod{19}.$$The quadratic residues mod 19 are $0, 1, 4, 9, 16, 6, 17, 11, 7, 5$. We can see that adding $0, -1, 1$ to any of these will not yield $14$, so it is impossible. It is funny how each individual equation is unnecessary as the problem can be solved using just the sum.

Adding the two equations and completing the square, we have $$(x^3+y+1)^2+z^9=147^{157}+157^{147}+1.$$We take mod 19. By FLT, the right side becomes $14^{13}+5^3+1$, and we can compute $14^{13}\equiv 14^8\cdot 14^4\cdot 14\equiv14\cdot-2\cdot4\equiv 2,$ so $14^{13}+5^3+1\equiv 14\pmod 19.$ Therefore, we then have $$(x^3+y+1)^2+z^9\equiv 14\pmod {19}.$$Since $(z^9)^2\equiv 0,1\pmod {19}$, we have $z^9\equiv -1,0,1\pmod {19}$. However, neither of $13,14,15$ are quadratic residues mod 19, so there are no solutions and we are done.

Her: He must be cheating on me Meanwhile me: Fakesolving this problem for 4 times straight and then finally making up my mind to case bash 18 total cases and literally compute thousands of modulos for 4 straight hours here is my solution that literally bases our every possible modulo all in a single problem you might have ever seen before Firstly manipulate the equations to get, $(x^3+1)(x^3+y)=147^{157}$ and $(y+1)(x^3+y)+z^9=157^{147}$. We have $\boxed{x^9\equiv\{0,\pm 1\}\pmod{27}}$ and $\boxed{x^9\equiv\{0,\pm 1\}\pmod{19}}$, which can be verified as $\varphi(27)=\varphi(19)=18$. Also, $\boxed{x^3\in\{0,1,7,8,11,12,18\}\pmod{19}}$ and $\boxed{x^3\in\{0,1,8,10,17,19,26\}\pmod{27}}$, which can be verified by case bashing each and every $\mod 19$ and $\mod 27$. Now if $27\mid x^3+y$, then taking the second equation $\mod{27}$ gives contradiction. Also notice that $\operatorname{ord}_{19}(3)=18$, $\operatorname{ord}_{19}(7)=3$ and $\operatorname{ord}_{27}(7)=3$. Now we case bash for $x^3+y$, with the powers of $3$ in each one of them and powers of $7$ taken $\mod 3$. Apart from this, we also have, \begin{align*} 147^{157}&\equiv 2\pmod{19}\\ 157^{147}&\equiv 11\pmod{19}\\ 147^{157}&\equiv 0\pmod{27}\\ 157^{147}&\equiv 10\pmod{27}. \end{align*} $\textbf{\underline{Case 1.1.1}: }x^3+y=1\cdot 7^{2\cdot 157-3i}$ and $x^3+1=3^{157}\cdot 7^{3i}$. Gives $x^3\equiv 7\pmod{19}$ and $y\equiv 4\pmod{19}$. But this contradicts $z^9\pmod{19}$. $\textbf{\underline{Case 1.1.2}: }x^3+y=3\cdot 7^{2\cdot 157-3i}$ and $x^3+1=3^{156}\cdot 7^{3i}$. Gives $x^3\equiv 8\pmod{19}$ and $y\equiv 6\pmod{19}$. But this contradicts $z^9\pmod{19}$. $\textbf{\underline{Case 1.1.3}: }x^3+y=9\cdot 7^{2\cdot 157-3i}$ and $x^3+1=3^{155}\cdot 7^{3i}$. Second one fails $\mod 19$. $\textbf{\underline{Case 1.2.1}: }x^3+y=1\cdot 7^{2\cdot 157-3i-1}$ and $x^3+1=3^{157}\cdot 7^{3i+1}$. Second one fails $\mod 19$. $\textbf{\underline{Case 1.2.2}: }x^3+y=3\cdot 7^{2\cdot 157-3i-1}$ and $x^3+1=3^{156}\cdot 7^{3i+1}$. Second one fails $\mod 19$. $\textbf{\underline{Case 1.2.3}: }x^3+y=9\cdot 7^{2\cdot 157-3i-1}$ and $x^3+1=3^{155}\cdot 7^{3i+1}$. Gives $x^3\equiv 1\pmod{19}$ and $y\equiv 5\pmod{19}$. But this contradicts $z^9\pmod{19}$. $\textbf{\textcolor{blue}{\underline{Case 1.3.1}:} }x^3+y=1\cdot 7^{2\cdot 157-3i-2}$ and $x^3+1=3^{157}\cdot 7^{3i+2}$. Gives $x^3\equiv 11\pmod{19}$ and $y\equiv 9\pmod{19}$. This works. $\textbf{\underline{Case 1.3.2}: }x^3+y=3\cdot 7^{2\cdot 157-3i-2}$ and $x^3+1=3^{156}\cdot 7^{3i+2}$. Second one fails $\mod 19$. $\textbf{\underline{Case 1.3.3}: }x^3+y=9\cdot 7^{2\cdot 157-3i-2}$ and $x^3+1=3^{155}\cdot 7^{3i+2}$. Second one fails $\mod 19$. $\textbf{\underline{Case 2.1.1}: }x^3+y=-1\cdot 7^{2\cdot 157-3i}$ and $x^3+1=-3^{157}\cdot 7^{3i}$. Second one fails $\mod 19$. $\textbf{\underline{Case 2.1.2}: }x^3+y=-3\cdot 7^{2\cdot 157-3i}$ and $x^3+1=-3^{156}\cdot 7^{3i}$. Second one fails $\mod 19$. $\textbf{\underline{Case 2.1.3}: }x^3+y=-9\cdot 7^{2\cdot 157-3i}$ and $x^3+1=-3^{155}\cdot 7^{3i}$. Second one fails $\mod 19$. $\textbf{\underline{Case 2.2.1}: }x^3+y=-1\cdot 7^{2\cdot 157-3i-1}$ and $x^3+1=-3^{157}\cdot 7^{3i+1}$. Gives $x^3\equiv 0\pmod{19}$ and $y\equiv 12\pmod{19}$. But this contradicts $z^9\pmod{19}$. $\textbf{\underline{Case 2.2.2}: }x^3+y=-3\cdot 7^{2\cdot 157-3i-1}$ and $x^3+1=-3^{156}\cdot 7^{3i+1}$. Gives $x^3\equiv 12\pmod{19}$ and $y\equiv 5\pmod{19}$. But this contradicts $z^9\pmod{19}$. $\textbf{\underline{Case 2.2.3}: }x^3+y=-9\cdot 7^{2\cdot 157-3i-1}$ and $x^3+1=-3^{155}\cdot 7^{3i+1}$. Second one fails $\mod 19$. $\textbf{\underline{Case 2.3.1}: }x^3+y=-1\cdot 7^{2\cdot 157-3i-2}$ and $x^3+1=-3^{157}\cdot 7^{3i+2}$. Second one fails $\mod 19$. $\textbf{\underline{Case 2.3.2}: }x^3+y=-3\cdot 7^{2\cdot 157-3i-2}$ and $x^3+1=-3^{156}\cdot 7^{3i+2}$. Second one fails $\mod 19$. $\textbf{\underline{Case 2.3.3}: }x^3+y=-9\cdot 7^{2\cdot 157-3i-2}$ and $x^3+1=-3^{155}\cdot 7^{3i+2}$. Second one fails $\mod 19$. We just have $\textbf{\textcolor{blue}{\underline{Case 1.3.1}:} }x^3+y=1\cdot 7^{2\cdot 157-3i-2}$ and $x^3+1=3^{157}\cdot 7^{3i+2}$ to check . Now for this case, we get $x^3\equiv -1\pmod{27}$ and $y\equiv 2\pmod{27}$. Now put this in the initial second equation and we now get a contradiction $z^9\pmod{27}$ YAYY!!

We can get $\begin{cases}147^{157}=(x^3+1)(x^3+y)\\157^{147}=(y+1)(x^3+y)+z^9\end{cases}$. Add up the two equations, we have $(x^3+y)(x^3+y+2)+z^9=147^{157}+157^{147}$. Let $n=x^3+y+1$, then $n^2+z^9=147^{157}+157^{147}+1\equiv 14\pmod {19}$. However, as $z^9\equiv\pm 1\pmod{19}$, $n^2\equiv 0,1,4,9,16,6,17,11,7,5\pmod {19}$, it is not true.$\blacksquare$

The key is to take mod $19$. Adding the two equations, $$(x^3+y+1)^2+z^9 \equiv 14 \pmod {19}.$$Now as $z^9 \equiv -1, 0, 1 \pmod {19}$, we must have $(x^3+y+1)^2 \equiv 13, 14, 15 \pmod {19}$. But none of these are quadratic residues.

Add the equations, we see that it can be factored as (x^3 +y +1)^2 + z^9 = 147^157 +157^147 +1. Note that z^9 In mod 19 is either 1 or -1 or 0. So this means that (x^3 +y+1)^2 is 13, 14, or 15 mod 19. Non of this are quadratic residues so there are 0 solutions.

evan why rate 10 MOHS Adding the equations, $$(x^3+y+1)^2+z^9 \equiv 14 \pmod {19};\quad z^9 \equiv\{ -1, 0, 1 \}\implies (x^3+y+1)^2 \equiv \{13, 14, 15\} \pmod {19},$$but none of these are quadratic residues, contradiction, so there are no solutions.

Note that by factoring, we have that \[(x^3+1)(x^3+y)=147^{157}\]\[(y+1)(x^3+y)+z^9=157^{147}.\]Since the latter is odd and $x$ and $y$ are integers, we have that $x$ must be even (because $x^3+1$ must be odd), meaning that $x^3\equiv 0 \mod 8$. Since $147^{157}\equiv3^{157}\equiv3\mod8$, we have that \[(x^3+1)(x^3+y)=147^{157}\equiv x^3+y \equiv 1+y \equiv 3 \mod 8 \iff y\equiv 2\mod8.\]Finally, note that $157^{147}\equiv5^{157}\equiv 5\mod 8$. Since both $y+1$ and $x^3+y$ are $3$ mod $8$, we have that $(y+1)(x^3+y)$ is $1$ mod $8$. Since \[(y+1)(x^3+y)+z^9=157^{147} \iff (y+1)(x^3+y)+z^9\equiv5 \mod8 \iff 1+z^9\equiv 5\mod8,\]this gives us that $z^9\equiv 4\mod8$, which is impossible. This is because the equation implies that $z$ is even, however, if $z$ was even, that would mean that $z^9$ would be $0$ mod $8$, a contradiction. Therefore there are no integer solutions to this system of equations, finishing the problem.

Factoring and adding the two equations, we get \[(x^3+y+1)^2 = 147^{157}+157^{147}+1-z^9.\] The use for modulo 19 becomes apparent, as $z^9 \pmod{19} \equiv \{-1,0,1\}$. Therefore the RHS is equivalent to $\{13,14,15\} \text{ mod } 19$, none of which quadratic residues modulo 19. $\blacksquare$

Adding the two equations and factoring, we get \[(x^3+y+1)^2 + z^9 = 147^{157}+157^{147}+1\]\[\implies (x^3+y^2+1)^2 \equiv \{13,14,15\} \pmod{19}.\] This results in no integer solutions as none of them are quadratic nonresidues modulo $19$. $\square$

Add both equations and add $1$ to both sides to get $x^6 + 2x^3y + 2x^3 + y^2 + 2y + 1 + z^9 = 147^{157} + 157^{147} + 1$. Factoring gets us $\newline$ \[(x^3 + y + 1)^2 + z^9 = 147^{157} + 157^{147} + 1\]Taking the expression modulo $19$ results in \[(x^3 + y + 1)^2 + z^9 \equiv 14\pmod{19}\]Notice that All QRs of $z^k$ modulo $2k + 1$ are $2k$, $0$, and $1$. So then we have \[(x^3 + y + 1)^2 + z^9 \equiv \{13, 14, 15\}\pmod{19}\]None of these being a QR, so there are no solutions.

Really profound problem. Summing both equations and manipulating we end up with $$(x^3+y+1)^2+z^9-1 = 157^{147}+147^{157}$$ Now consider $\mod{19}$, in which case we can restrict $z^9$ to $\{1,0,-1\}$. Then we can calculate by hand the QR of $\mod{19}$ which is $\{0,1,4,9,6,17,30,7,5\}$. Then we can calculate that $157^{147} \equiv 30 \equiv 11 \mod{19}$ and $147^{157} \equiv 17^3(-5) \equiv 40 \equiv 2 \mod{19}$. So we need $\{1,0,-1\}+\{0,1,4,9,6,17,30,7,5\} \equiv 14 \mod{19}$ which is clearly not possible.

Motivation Add and subtract with equation. Note that when adding, you get a format for perfect squares like $x^6+2x^3$ and $y^2+2y$, but you also have $2x^3y$ left over. Thus, can we combine this into one expression? Yes! Let's try $(x^3+y+1)^2$. Now, what modulo do we take? After testing classics such as $3,7,9$, let's pick one such that $p-1$ is divisible by $9$ by FLT. Let's pick base $19$! Solution Continuing, \begin{align*} (x^3+y+1)^2 +z^9 \equiv 14 \pmod{19} \\ \end{align*} Now note that $z^9 \equiv 0,1,-1 \pmod{19}$. Thus, we get $$(x^3+y+1)^2 +z^9 \equiv 14 \pmod{19}$$which is not possible.

what is this Adding the equations and adding $1$ to both sides gives $$z^9+x^6+2x^3y+2x^3+y^2+2y+1=147^{157}+157^{147}+1$$$$\implies z^9+(x^3+y+1)^2=147^{157}+157^{147},$$but $\pmod {19}$ there are no solutions.

Adding the equations yields $(x^3+y+1)^2+z^9 = 147^{157}+157^{147}+1.$ However, $z^9 \equiv \{-1, 0, 1\} \pmod{19},$ so it follows that $$(x^3+y+1)^2 \equiv \{13, 14, 15 \} \pmod{19}.$$However, none of these are quadratic residues mod $19,$ so we are done. QED