IMO 1991 SL 23

Origin: IND

Problem

Let f and g be two integer-valued functions defined on the set of all integers such that (a) f(m + f(f(n))) = −f(f(m + 1) −n for all integers m and n; (b) g is a polynomial function with integer coefficients and g(n) = g(f(n)) for all integers n. Determine f(1991) and the most general form of g.

Solution

From (i), replacing m by f(f(m)), we get f( f(f(m)) + f(f(n)) ) = −f(f( f(f(m)) + 1)) −n; analogously f( f(f(n)) + f(f(m)) ) = −f(f( f(f(n)) + 1)) −m. From these relations we get f(f(f(f(m))+1))−f(f(f(f(n))+1)) = m−n. Again from (i), f(f( f(f(m)) + 1)) = f(−m −f(f(2)) ) and f(f( f(f(n)) + 1)) = f(−n −f(f(2)) ). Setting f(f(2)) = k we obtain f(−m −k) −f(−n −k) = m −n for all integers m, n. This implies f(m) = f(0) −m. Then also f(f(m)) = m, and using this in (i) we finally get f(n) = −n −1 for all integers n. Particularly f(1991) = −1992. From (ii) we obtain g(n) = g(−n −1) for all integers n. Since g is a polynomial, it must also satisfy g(x) = g(−x −1) for all real x. Let us now express g as a polynomial on x + 1/2: g(x) = h(x + 1/2). Then h satisfies h(x + 1/2) = h(−x −1/2), i.e. h(y) = h(−y), hence it is a polynomial in y2; thus g is a polynomial in (x + 1/2)2 = x2 + x + 1/4. Hence g(n) = p(n2 + n) (for some polynomial p) is the most general form of g.