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 2 of 4: Use the other quadratic root
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A+C=kB,AC=B2−k,C=kB−A=B2−kAA+C=kB,\qquad AC=B^2-k,\qquad C=kB-A=\frac{B^2-k}{A}
Detailed analysis

Regard A2+B2=k(AB+1)A^2+B^2=k(AB+1) as a quadratic in A, with B=bB=b and A=aA=a. Its other root C satisfies A+C=kBA+C=kB and AC=B2−kAC=B^2-k, hence C=kB−A=(B2−k)/AC=kB-A=(B^2-k)/A. In particular C is an integer, and C<BC<B because A≥BA\ge B and k>0k>0.