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Sauda,c~oes,
Tive problemas para enviar esta mensagem.
Mando-a em separado e junto com a outra
do assunto original em reply.
Este problema caiu no 61o Concurso Putnam.
Acho que corresponde ao ano 2000. Não me lembrava mais que o prof. Rousseau havia me
mandado a solução deste problema. Aí vai ela:
SIXTY-FIRST ANNUAL WILLIAM LOWELL PUTNAM
COMPETITION
{\bf Probem A2.} Prove that there exist infinitely many
integers
$n$ such that $n, n+1$, and $n+2$ are each the sum of two
squares.
Example: $0 = 0^2 + 0^2; 1 = 0^2 + 1^2; 2 = 1^2 +
1^2.
{\bf Solution.} For completeness, we first prove the
following
well-known fact.
{\sc Lemma.} If $m$ and $n$ are each the sum of two squares,
then
so is $mn$. More generally, if $m_1, m_2, \ldots, m_k$ are
each
the sum of two squares, then so is $m_1m_2\cdots
m_k$.
\begin{proof} Suppose $m = a^2 + b^2$ and $n = c^2 +
d^2$.
Then
\[
mn = (ac - bd)^2 + (ad + bc)^2.
\]
Having proved the result for $k = 2$, the general result
follows
by an obvious use of mathematical induction.
\end{proof}
Suppose $k$ and $k+1$ are each the sum of two squares, and
set
$n = (2k+1)^2 - 1 = 4k(k+1)$.
Since $4 = 2^2 + 0^2$, $k$, and $k+1$ are
each the sum of two squares, the lemma shows
that $n$ is the sum of two squares. Also
$n+1 = (2k+1)^2 + 0^2$ and
$n+2 = (2k+1)^2 + 1^2$ are each the
sum of two squares. Clearly $n > k$.
The fact that there are infinitely many triples $(n,n+1,n+2)$ where each member is
the sum of two squares follows inductively.
[]'s
Luís
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