If $a,b,c,d$ are natural numbers, prove that $6(6a^2+3b^2+c^2)=5d^2$ has no solutions.
I did some simplifying and got that I need to prove $2a^2+b^2+3x^2=10k^2$ has no integral solutions where $a,b,x,k$ are natural numbers. But I'm not able to proceed any further.
This question is very famous, and you should have found some sort of solution if you had searched well enough. Still here is a solution using Fermat's Principle of Infinite Descent:
I am proceeding from $2a^2+b^2+3x^2=10k^2.$ Taking$\mod 2$ on both sides we can see that, $2 \mid b^2+3x^2.$ Thus, $b$ and $x$ have the same parity.
Case I: $b$ and $x$ are odd.
Subcase I: $a$ is even.
Take $\mod 8$ on both sides. We get, LHS$\equiv 0+1+3 =4\pmod {8}$, and RHS$\equiv 0,2\pmod 8$. Thus this case is discarded.
Subcase II: $a$ is odd.
Take $\mod 8$ on both sides, yielding: LHS$\equiv 2+1+3=6\pmod8$, and RHS$\equiv 0,2\pmod8$. Thus, this case is also discarded.
Case II: $b$ and $x$ are even.
Let, $b=2B$ and $x=2X$. The equation then becomes: $$a^2+2B^2+6X^2=5k^2.$$
Taking $\mod 2$ on both sides, we get: $2 \mid 5k^2-a^2.$
So, $a$ and $k$ are of same parity. Proceeding further in the same way as before, we can show that $a$ and $k$ are both even. Let $a=2A, k=2K$. The equation then becomes: $$2A^2+B^2+3X^2=10K^2.$$
This is very like the first equation we had started with, just the variables are reduced. So, by Fermat's Principle of Infinite Descent, we get that no positive integer solution of the original equation exists. We can show that even negative solutions do not exists.(just negate each variable and see that the equation remains exactly the same.) Indeed, the only integer solution is $(0,0,0,0).$
