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 $49$

Answer: E

Concepts:triangular square numbers Pell equation recurrence

Difficulty rating: 2450

Small Hint:

Rewrite $\frac{K(K+1)}{2}=N^2$ as $(2K+1)^2-8N^2=1$

Big Hint:

Generate successive positive solutions by multiplying $(2K+1)+N\sqrt8$ by $3+\sqrt8,$ stopping when $N\ge100$

Solution:

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.

1956 AMC 12 Problem 36

Answer: E

Difficulty rating: 2450

Small Hint:

Big Hint:

Solution:

Problem 36 in Other Years