Spring Mathematical Tournament

We shall prove that $a=b=c=d=0$. It is easy to see that if one of the numbers $a,b,c,d$ equals $0$, then the others equal $0$, too. Indeed, if $b=0$, $a\ne 0$, then $\sqrt{3}=\pm \frac{c}{a^2}$ is a rational number, a contradiction. The case $a=0$, $b\ne 0$ is impossible by the same reasoning. Let $c=0$. Then $n=-b>0$ and $3a^4=2n^3$, $3a^6\ge n^5$. Hence $2a^2\ge n^2$ and therefore $2n^3=3a^4\ge \frac{3n^4}{4}$. Then $n\le \frac{8}{3}$, i.e. $n=1$ or $n=2$, a contradiction. Analogously $d\ne 0$.

Let now $a,b,c,d\ne 0$. Adding both equations and using that $3$ divides $b^5+2b^3=b^3(b-1)(b+1)+3b^3$, we get that $3|c^2+d^2$, i.e. $3|c,d$. Hence $3|a,b$. Let $a=3^{\alpha }{a}_1$, $b=3^{\beta }{b}_1$, $c=3^{\gamma }{c}_1$, $d=3^{\delta }{d}_1$, where $\alpha ,\beta ,\gamma ,\delta \ge 1$ and $3\nmid a_1,b_1,c_1,d_1$. Then the system can be written in the form $$(1)|\begin{array}{l}{3}^{4\alpha +1}{a}_1^4+3^{3\beta }2b_1^3=3^{2\gamma }{c}_1^2\\ 3^{6\alpha +1}{a}_1^6+3^{5\beta }{b}_1^5=3^{2\delta }{d}_1^2\end{array}$$ We shall use the following trivial fact. If $3^{k}p+3^{l}q=3^{m}r$ and $3\nmid p,q,r$, then at least two of the numbers $k,l,m$ are equal. This and (1) imply that $4\alpha +1=3\beta$ or $3\beta =2\gamma$, and $6\alpha +1=5\beta$ or $5\beta =2\delta$. Then it is easy to see that $3\beta =2\gamma$ and $5\beta =2\delta$. Now (1) is equivalent to $$|\begin{array}{l}{3}^{4\alpha +1-2\gamma }{a}_1^4+2b_1^3=c_1^2\\ 3^{6\alpha +1-2\delta }{a}_1^6+b_1^5=d_1^2\end{array}$$ Since $4\alpha +1-2\gamma >0$ and $6\alpha +1-2\delta >0$, adding the last two equations, we conclude as above that $3|c_1,d_1$, a contradiction.