Olimpiadi di Matematica

Solution:

There are 6 pairs of triangular numbers that satisfy the required condition. The problem is equivalent to finding the pairs of positive integers $(n, m)$ such that
$$\frac{n(n+1)}{2}-\frac{m(m+1)}{2}=2007.$$
Indeed, since different values of $n$ determine different values of $\frac{n(n+1)}{2}$, determining the triangular numbers $a$ and $b$ is equivalent to determining the values of $n$ and $m$ that generate them.
Clearing denominators and factoring, the equation becomes
$$(n-m)(n+m+1)=2 \cdot 2007=2 \cdot 3^{2} \cdot 223.$$
Since $n>0, m>0$ and hence $n+m+1>0$, $n-m$ must also be positive. The equation then represents a way of writing $2 \cdot 2007$ as a product of two positive integers. But the only ways of splitting $2 \cdot 3^{2} \cdot 223$ as a product of two positive integers are obtained by "distributing" its prime factors in all possible ways between the two integers, that is, explicitly
- 1 and $2 \cdot 3 \cdot 3 \cdot 223$;
- 2 and $3 \cdot 3 \cdot 223$;
- 3 and $2 \cdot 3 \cdot 223$;
- $2 \cdot 3$ and $3 \cdot 223$;
- $3 \cdot 3$ and $2 \cdot 223$;
- $2 \cdot 3 \cdot 3$ and 223
Since $(n+m+1)-(n-m)=2 m+1>0$, $n-m$ is always the smaller of the two factors of the pair. Letting $u=n+m+1$ and $v=n-m$, we must then be in one of the following cases:
- $v=1, u=2 \cdot 3 \cdot 3 \cdot 223$;
- $v=2, u=3 \cdot 3 \cdot 223$;
- $v=3, u=2 \cdot 3 \cdot 223$;
- $v=2 \cdot 3, u=3 \cdot 223$;
- $v=3 \cdot 3, u=2 \cdot 223$;
- $v=2 \cdot 3 \cdot 3, u=223$.

Now solving the system
$$\left\{ \begin{array}{l} n+m+1=u \\ n-m=v \end{array} \right.$$
in the unknowns $n$ and $m$ we obtain
$$\left\{ \begin{array}{l} n=\frac{u+v-1}{2} \\ m=\frac{u-v-1}{2} \end{array} \right.$$
Note that substituting into the equations the six values of $(u, v)$ corresponding to the six possible cases, one finds each time a pair of acceptable values $(n, m)$ (that is, integer and positive: indeed we always have $u-v>1$, and, since $2 \cdot 2007$ has only one factor of 2, in all cases exactly one of $u$ and $v$ is even, and therefore $u+v-1$ and $u-v-1$ are always even). For all six of these pairs, by the way they were obtained, we have $(n+m+1)(n-m)=2 \cdot 2007$, and therefore all of them are solutions of the original equation. Moreover the six cases presented exhaust all the possibilities, so there are no other solutions. Explicitly the solutions are:
- $n=2007, m=2006$;
- $n=1004, m=1002$;
- $n=670, m=667$;
- $n=337, m=331$;
- $n=227, m=218$;
- $n=120, m=102$
(note however that the exercise does not ask to explicitly determine the solutions, but only to say how many there are).