MathLabs

Problem 2

Prove that the equation 6(6a2+3b2+c2)=5n26(6a^2+3b^2+c^2)=5n^2 has no integer solutions except a=b=c=n=0a=b=c=n=0.
Step 2 of 5: Force c to be divisible by 3
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3∣c,c=3c0,2a2+b2+3c02=10n023\mid c,\quad c=3c_0,\quad 2a^2+b^2+3c_0^2=10n_0^2
Detailed analysis

Reducing 6a2+3b2+c2=30n026a^2+3b^2+c^2=30n_0^2 modulo 33 gives c2≡0(mod3)c^2\equiv0\pmod3, so c=3c0c=3c_0. Dividing by 33 yields 2a2+b2+3c02=10n022a^2+b^2+3c_0^2=10n_0^2.