Prove that if a²+b² is a multiple of ab+1, for positive integer a and b, then (a²+b²)/(ab+1) is a perfect square.

(In reply to

Puzzle Solution by K Sengupta)

In consonance with the methodology in terms of the previous post, we take (a^2+b^2)/(ab+1) = p, so that each of a, b and e are positive integers.

Now, fixing p, we consider positive integers C and D such that

C^2 + D^2 = p(CD + 1)......(#).

We also consider Min(C, D) as small as possible.

Without loss of generality, let us assume that C> =D.

From (#), C^2 -pCD +(D^2-p)=0

Regarding this as a quadratic for C, we observe that the other root E (say) satisfies C+E = pD and CE = D^2-p.

Now, CE = D^2-p gives E = D^2/C - p/C < D, so that:

Min(E, D) < Min (C, D). This is a contradiction.

Consequently, E cannot be a positive integer.

Now,

(C+1)(E+1)

= C+E+CE+1

= D^2 +(D-1)p+1

> 0

So, either both (C+1, E+1) must exceed 0, or both (C+1, E+1) must be

less than 0.

So, E+1> 0

Or, E> -1

But, E = pD-C is an integer

Since E is not a positive integer, this is possible iff E =0.

From CE = D^2-p, it thus follows that p=D^2.

Consequently, the quotient (a^2+b^2)/(ab+1) is always a perfect square.

*Edited on ***March 26, 2007, 4:21 am**