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 5 of 5: Descend to a smaller solution
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2∣a,b,c,n⇒6(6(a/2)2+3(b/2)2+(c/2)2)=5(n/2)22\mid a,b,c,n\Rightarrow 6(6(a/2)^2+3(b/2)^2+(c/2)^2)=5(n/2)^2
Detailed analysis

We have shown that a,b,c,na,b,c,n are all even. Dividing the original equation by 44 gives the displayed equation for (a/2,b/2,c/2,n/2)(a/2,b/2,c/2,n/2), a solution with smaller absolute-value sum. This contradicts minimality unless all four variables were already zero.