1956 AMC 12 Problem 36

Question

If the sum

1+2+3+\cdots+K

is a perfect square

N^2

and if

N

is less than

100,

then the possible values for

K

are:

only

1

1

and

8

only

8

8

and

49

1,

8,

and

49

Rewrite

\frac{K(K+1)}{2}=N^2

as

(2K+1)^2-8N^2=1

Generate successive positive solutions by multiplying

(2K+1)+N\sqrt8

by

3+\sqrt8,

stopping when

N\ge100

The condition is

\frac{K(K+1)}2=N^2,

or

(2K+1)^2-8N^2=1.

The positive solutions of this Pell equation are generated from the fundamental solution

3+\sqrt8.

Their first pairs

(2K+1,N)

are

\begin{aligned} &(3,1),\ (17,6),\\ &(99,35),\ (577,204),\ldots \end{aligned}

Thus the values with

N\lt100

are

K=1,8,49.

Therefore, the correct answer is E.

Problem 36 in other years