Problem 2

Prove that the equation

6(6a^2+3b^2+c^2)=5n^2

has no integer solutions except

a=b=c=n=0

Step 4 of 5: The second modulo-8 argument makes a and n0 even

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a^2+2b_0^2+6c_1^2=5n_0^2\pmod{8}\Rightarrow 2\mid a,\ 2\mid n_0

Detailed analysis

Substitution and division by $2$ give $a^2+2b_0^2+6c_1^2=5n_0^2$ . Modulo $2$ , $a,n_0$ have the same parity. If both were odd, modulo $4$ forces $b_0,c_1$ to have the same parity; if both are odd, modulo $8$ gives a contradiction, while if both are even the equation reads $1\equiv5\pmod8$ , also a contradiction. Hence both $a$ and $n_0$ are even.