**POLYA002: Integer pairs (x,y) for which (x^2+y^2)/(1+pxy) is an integer
**(proposed
by Jean-Pierre Ehrmann, 11/7/01).

Given positive integers p, q satisfying pq > 1, find the set of integer
pairs {x,y} such that x^2+y^2 = q(pxy+1).

**Discussion.** **[PY, 11/9/01]:** After typing
this up, by coincidence, I ran into this story on p.127 of A.Engel's ``Problem-Solving
Strategies'':

The problem [IMO,1988: If a,b, q = (a^2+b^2)/(ab+1) are
integers, then q is a perfect square] was submitted in 1988 by the FRG.
Nobody of the six members of the Australian problem committee could solve
it. Two of the members were Georges [sic] Szekeres and his wife,
both famous problem solvers and problem creators. Since it was a number
theoretic problem it was sent to the four most renowned Australian number
theorists. They were asked to work on it for six hours. None of them could
solve it in this time. The problem committee submitted it to the jury of
the XXIX IMO marked with a double asterisk, which meant a superhard problem,
possibly too hard to pose. After a long discussion, the jury finally had
the courage to choose it as the last problem of the competition. Eleven
students gave perfect solutions.

Now I recall having heard the same story from Hans-Dietrich Gronau 2 years ago.

**[JPE, 11/9/01]: ** I don't understand why this
exercice seems so difficult. May be because a number theorist knows too
much things to do it. The only thing to know is the sum and the product
of the roots of a quadratic equation. If (a,b) is a solution with a>= b
> 0, then the roots of x^2 - qbx + b^2-q = 0 are a and c=qb-a=(b^2-q)/a.
Hence (b,c) is another solution and b>c>=0. Hence, if we choose b minimal,
we get c=0 and q=b^2.

**[PY, 11/9/01]:** Let me try to understand you. For
a positive integer q, let write B_q for the set of positive integers b
for which there exist a (>= b) such that a^2 - qba + b^2-q = 0. If B_q
is nonempty, let c the smallest integer in be smallest element. Necessarily,
c^2 = q. This settles the IMO question.

But one can go further. With q = c^2, the equation becomes
a^2 - c^3a = 0, and a = c^3. [Check: a^2+c^2/(ac+1) = c^2(c^4+1)/(c^4+1)
= c^2 = q].

Your method of descent actually leads to a description
of the set of integers (a,b) satisfying (a^2+b^2)/(1+ab) = integer c^2.
These pairs are the consecutive terms of the sequence (u_n) defined
recursively by

u_(n+2) = c^2.u_(n+1) - u_n, u_0 = c, u_1 = c^3.

Explicitly, the beginning terms of the sequence are c,
c^3, c^5-c, c^7-2c^3, c^9-3c^5+c, c^11-4c^7+3c^3, ...

For c = 2, this is the sequence 2, 8, 30, 112, 418, 1560,
5822, 21728, 81090, ...

For c = 3, this is 3, 27, 240, 2133, 18957, 168480, 1497363,
...

Now I understand why you asked the question. Given p
and q with pq>1, there are integers (x,y) satisfying the quadratic equation
(x^2+y^2)/(1+pxy) = q if and only if q = c^2 for an integer c. The pairs
of (x,y) satisfying this condition are the consecutive terms of the sequence

U_(n+2) = pc^2.U_{n+1} - U_n, U_0 = c, U_1 = pc^3.

The beginning terms are c, pc^3, p^2c^5-c, p^3c^7-2pc^3,
p^4c^9-3p^2c^5+c, p^5c^11-4p^3c^7+3pc^3, ...

For p = 3, q = 2, this is the sequence 2, 24, 286, 40610,
483912, 5766334, ...

For p=2, q=3, this is 3, 54, 969, 17388, 312015, 5598882,
...

Very interesting. Thank you for a good problem.

**[JPE, 11/9/01]: ** That was exactly my conclusion
and my way to get the result. Note that u[n] = sum((-1)^k C(n-k,k) p^(n-2k)
c^(2n-4k+1), 0<=k<=n/2) and that u[n] is closely related to the Chebychev
polynomial P[n] such as sin(n+1)x = P[n](2 cos x) sin x or sinh(n+1)x =
P[n](2 cosh x) sinh x. In fact, u[n] = cP[n](pc^2). Another interesting
conclusion is that c is the gcd of x,y for any pair of solutions.

**[FvL: 11/28/01]:** Perhaps it is interesting to extend
this problem to **triangular numbers** t(n)=n(n+1)/2.

In a way very similar to the IMO problem, we can prove
that if, a, b and q =( t(a) + t(b) )/( 1 + ab ) are integers, then N is
a triangular number.

We get ( a(a+1) + b(b+1) ) / ( 2 + 2ab ) = q rewritten
to a^2 + (1-2bq)a + b^2 + b - 2q = 0.

The roots of x^2 + (1-2bq)x + b^2 + b - 2q = 0 are x=a
and x=2bq-a-1=(b^2+b-2q)/a=:c. When a>b and thus a>=b+1 we can show that
c<b.

Etc. The sequence yielding pairs u_n and u_{n+1} such
that q=t(m) are given by u_n = 2t(m)u_{n-1}-u_{n-2}-1, u_0 = m, u_1 = m^3
+ m^2 - 1.

So q=3 gives 2, 11, 63, 366, 2132, 12425, 72417, ...

q=6 gives 3, 35, 416, 4956, 59055, 703703, ...

I have not gone into denominators (1+pab). All does not
work for n-gonal numbers with n>4 in stead of the triangular or square
numbers, this does not work. Then the second root of the quadratic equation
is not necessarily an integer. For n=5 I made a small table indicating
that indeed "q" is not pentagonal.

**[FvL: 11/29/01]: **Perhaps it is interesting to generalize
JPE's original question in another way (than the triangular numbers of
yesterday): In stead of putting p in the denominator we could also put
p into the numerator as follows:** (x^2 + pxy + y^2)/(1+xy) = q.**

An interesting example is when we take p=2 and q=9, then
we find as solution sequence: 3, 21, 144, 987, 6765, 46368, 317811, ...
which are **Fibonacci(4n)**.

Since 1+xy must be a square as good as the numerator
and q are, we know that sqrt(u(n)u(n+1)+1) gives an integersequence, which
is indeed 1, 8, 55, 377, 2584, 17711, 121393, ... which are **Fibonacci(4n+2).**

Taking q = c^2 the recurrence relation is u_n = (c^2-p)u_{n-1)
- u_{n-2}, u_0 = c, u_1 = c^3 - pc.

**[FvL: 12/3/01]: ** In my last contribution to
this discussion I have been imprecise.

We can only apply the IMO-method to (x^2 + pxy + y^2)/(1+xy)
= q when the numerator is pos.def. This is only the case for -2 < p
< 2, so the example, (x+y)^2/(1+xy)=9, I gave was not solved completely.
In fact there is a second sequence with solutions, being Fibonacci(4n+2).

**Bibliography. **[1] A.Engel, *Problem -Solving
Strategies*, Springer, 1998.

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