Let-m-be-the-smallest-odd-positive-integer-for-which-1-2-m-is-a-square-of-an-integer-and-let-n-be-the-smallest-even-positive-integer-for-which-1-2-n-is-a-square-of-an-integer-What

Question Number 19696 by Tinkutara last updated on 14/Aug/17

Let m be the smallest odd positive

integer for which 1 + 2 + \dots + m is a

square of an integer and let n be the

smallest even positive integer for

which 1 + 2 + \dots + n is a square of an

integer . What is the value of m + n ?

Answered by Rasheed.Sindhi last updated on 15/Aug/17

Case - 1 : m \in O

1 + 2 + \dots m = \frac{m (m + 1)}{2}

Let m = 2 k + 1

\frac{m (m + 1)}{2} = \frac{(2 k + 1) (2 k + 2)}{2} = a^{2} (s a y)

\frac{{4 k}^{2} + 4 k + 2 k + 2}{2} = a^{2}

{2 k}^{2} + 3 k + 1 - a^{2}

k = \frac{- 3 \pm \sqrt{9 - 4 (2) (1 - a^{2})}}{4}

k = \frac{- 3 \pm \sqrt{1 + {8 a}^{2})}}{4}

a = 1 \Rightarrow k = \frac{- 3 \pm 3}{2} = 0, - 3

k = 0 \Rightarrow m = 2 k + 1 = 1

m = 1

is the smallest positive

integer for which 1 + 2 + \dots + m

is perfect square .

k = - 3 \Rightarrow m = 2 (- 3) + 1 = - 5 (- ve)

So - 5 is discardable .

a = 6 \Rightarrow \sqrt{1 + 8 {(6)}^{2}} = 17

k = \frac{- 3 \pm 17}{4} = \frac{7}{2}, - 5

Both are discardable .

Case - 2 : n \in E

1 + 2 + \dots + n = b^{2}

\frac{n (n + 1)}{2} = b^{2}

Let n = 2 k

\frac{2 k (2 k + 1)}{2} = b^{2}

{2 k}^{2} + k - b^{2} = 0

k = \frac{- 1 \pm \sqrt{1 - 4 (2) (- b^{2})}}{4}

k = \frac{- 1 \pm \sqrt{1 + {8 b}^{2}}}{4}

b = 1 \Rightarrow k = \frac{- 1 \pm 3}{4} = - 1, \frac{1}{2}

Both are discardable .

b = 6 \Rightarrow k = \frac{- 1 \pm 17}{4} = - \frac{9}{2}, 4

- \frac{9}{2} is discardable .

k = 4 \Rightarrow n = 8

1 + 2 + \dots + 8 = \frac{8 \times 9}{2} = 36

n = 8

is the smallest positive

integer for which 1 + 2 + \dots + n

is perfect square .

Commented by Tinkutara last updated on 15/Aug/17

Thank you very much Sir!