Solve, given all unknowns are non-zero integers
$$\begin{align}a^2 &= b^3 + c^3 \\b^2 &= c^3 + d^3 \\c^2 &= d^3 + e^3\\d^2 &= e^3 + f^3\\e^2 &=f^3 + g^3 \\f^2 &= g^3 + a^3\end{align}$$
Any tips on solving this please?
I can solve a single equation $a^2=b^3+c^3$, but Im having trouble going the next step
Thanks
By examining this mod powers of $2$, you can at the least conclude some lower bounds on the absolute value of these numbers and some fairly large powers of $2$ that they must be divisible by. If you follow the argument below and see some clever pattern I am missing, then the below might be workable into an argument that no solutions exist.
Mod $2$, you have $x^n\equiv x$ for all $n$. So the system is: $$ \begin{align} a &\equiv b + c & b &\equiv c + d & c &\equiv d + e& d &\equiv e + f& e &\equiv f + g& f &\equiv g + a \end{align}$$
Summing all the equations gives $a+b+c+d+e+f\equiv a+b$, which reduces to $c+d\equiv e+f$. Adding to the fourth equation, $c\equiv0$. Then the first, second and third equations say $a\equiv b\equiv d\equiv e$. Then the fourth equation says $f\equiv 0$, so the fifth equation says $g$ is also equivalent to $a\equiv b\equiv d\equiv e$.
So $c$ and $f$ are even for sure. And either all other values are even or all other values are odd.
Mod $8$, any odd number squared is $1$. Any odd number cubed is itself. And any even number cubed is $0$. So if $a,b,d,e,g$ are all odd, then mod $8$ the system is:$$ \begin{align} 1 &\equiv b& 1 &\equiv d& c^2 &\equiv d + e& 1 &\equiv e& 1 &\equiv g& f^2 &\equiv g + a \end{align}$$
But this is an impossibility when you consider that the second, third, and fourth equations imply $c^2\equiv2$ which has no solutions mod $8$.
So all seven values are even. Mod $2^3$, the system is $$ \begin{align} a^2 &\equiv 0& b^2 &\equiv 0& c^2 &\equiv 0& d^2 &\equiv 0& e^2 &\equiv 0& f^2 &\equiv 0 \end{align}$$
which implies $a,b,c,d,e,f$ are all divisible by $2^2$. Note that the last two original equations then force $g$ to be divisible by $2^2$. Then mod $2^6$, the system is:$$ \begin{align} a^2 &\equiv 0& b^2 &\equiv 0& c^2 &\equiv 0& d^2 &\equiv 0& e^2 &\equiv 0& f^2 &\equiv 0 \end{align}$$
which implies $a,b,c,d,e,f$ are all divisible by $2^3$. Then mod $2^9$, the system is:$$ \begin{align} a^2 &\equiv 0& b^2 &\equiv 0& c^2 &\equiv 0& d^2 &\equiv 0& e^2 &\equiv g^3& f^2 &\equiv g^3 \end{align}$$
which implies $a,b,c,d,$ are all divisible by $2^5$. Note the original third equation says $e^3=c^2-d^3$ which is divisible by $2^{10}$, so $e$ is divisible by $2^4$. And then the original fourth equation says $f^3=d^2-e^3$, which is divisible by $2^{10}$. So $f$ is also divisible by $2^4$. And then the original fifth equation says $g^3=e^2-f^3$, divisible by $2^8$. So $g$ is divisible by $2^3$. Then $e^2=f^3+g^3$ is divisible by $2^9$, so $e$ is divisible by $2^5$. Then $g^3=e^2-f^3$ is divisible by $2^{10}$, so $g$ is divisible by $2^4$. The sixth original equation similarly gives that $f$ is divisible by $2^5$. In summary to this point, $a,b,c,d,e,f$ are all divisible by $2^5$ and $g$ is divisible by $2^4$.
OK, now mod $2^{15}$, the system is:$$ \begin{align} a^2 &\equiv 0& b^2 &\equiv 0& c^2 &\equiv 0& d^2 &\equiv 0& e^2 &\equiv g^3& f^2 &\equiv g^3 \end{align}$$
Note the growth in the power of $2$ in the modulus. And we have the same reduced system as a short while ago. We can immediately conclude $a,b,c,d$ are divisible by $2^8$. With similar pencil pushing as from the previous round, we can find higher powers of $2$ dividing $e,f$ as well. I'm too out of steam to type it up, but $e$ and $f$ can be proved to be divisible by $2^6$.
Then yet another round of this approach (mod $2^{18}$) leaves me with $a,b,c,d$ divisible by $2^9$, $e,f$ divisible by $2^6$, and $g$ divisible by $2^4$.
More and more of this approach keeps raising the powers of $2$ dividing the values. I've gotten up to $a,b$ divisible by $2^{21}$, $c,d$ by $2^{14}$, $e,f$ by $2^{9}$, $g$ by $2^{6}$ and I'm calling it quits even though I know it will go higher.
Either there is a pattern to capture and then make an inductive infinite ascent argument that no solutions exist; or at the worst if you keep up with this pencil-pushing you will know some large powers of $2$ that must divide each value.