Methods of Proof 3351 = 12 ,
2 =− 12 − 22 − 32 + 42 ,
3 =− 12 + 22 ,
4 =− 12 − 22 + 32.
The inductive step is “P (n)impliesP(n+ 4 )’’; it is based on the identity
m^2 −(m+ 1 )^2 −(m+ 2 )^2 +(m+ 3 )^2 = 4.Remark.This result has been generalized by J. Mitek, who proved that every integerk
can be represented in the formk=± 1 s± 2 s±···±msfor a suitable choice of signs,
wheresis a given integer≥2. The number of such representations is infinite.
(P. Erdos, J. Surányi) ̋
31.First, we show by induction onkthat the identity holds forn= 2 k. The base case is
contained in the statement of the problem. Assume that the property is true forn= 2 k
and let us prove it forn= 2 k+^1. We have
f(
x 1 +···+x 2 k+x 2 k+ 1 +···+x 2 k+ 1
2 k+^1)
=
f(x
1 +···+x 2 k
2 k)
+f(x
2 k+ 1 +···+x 2 k+^1
2 k)
2
=
f(x 1 )+···+f(x 2 k)
2 k +f(x 2 k+ 1 )+···+f(x 2 k+ 1 )
2 k
2
=f(x 1 )+···+f(x 2 k)+f(x 2 k+ 1 )+···+f(x 2 k+ 1 )
2 k+^1,
which completes the induction. Now we work backward, showing that if the identity
holds for somen, then it holds forn−1 as well. Consider the numbersx 1 ,x 2 ,...,xn− 1
andxn=x^1 +x^2 n+···+− 1 xn−^1. Using the hypothesis, we have
f(
x 1 +···+xn− 1 +x^1 +···+n− 1 xn−^1
n)
=
f(x 1 )+···+f(xn− 1 )+f(x 1 +···+xn− 1
n− 1)
n,
which is the same as
f(
x 1 +···+xn− 1
n− 1)
=
f(x 1 )+···+f(xn− 1 )
n+
1
nf(
x 1 +···+xn− 1
n− 1)
.
Moving the last term on the right to the other side gives
n− 1
n
f(
x 1 +x 2 +···+xn− 1
n− 1)
=
f(x 1 )+f(x 2 )+···+f(xn− 1 )
n