MathLabs

Problem 6

Let a,ba,b be positive integers such that ab+1ab+1 divides a2+b2a^2+b^2. Show that (a2+b2)/(ab+1)(a^2+b^2)/(ab+1) is the square of an integer.
Step 3 of 4: Minimality forces the other root to vanish
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(A+1)(C+1)=B2+(B−1)k+1>0⇒C>−1⇒C=0(A+1)(C+1)=B^2+(B-1)k+1>0\Rightarrow C>-1\Rightarrow C=0
Detailed analysis

If C were positive, (C,B) would be a smaller positive solution with the same k, contradicting minimality. Also A+C+AC+1=B2+(B−1)k+1>0A+C+AC+1=B^2+(B-1)k+1>0, so (A+1)(C+1)>0(A+1)(C+1)>0 and therefore C>−1C>-1. Since C is integral and cannot be positive, C=0.