Sum is a Perfect Square

Recall that $\sum_{k=1}^N k = N(N+1)/2$ . Suppose $N(N+1)/2 = M^2$ . Since $N, N+1$ cannot have any common factors greater than 1 (as they are relatively prime), either $N = 2y^2$ or $N+1 = 2y^2$ . In the first case, we obtain $M^2 = y^2(2y^2+1)$ or $x^2 = (M/y)^2 = 2y^2+1$ . Similarly, in the second case, we obtain $M^2 = (2y^2 - 1)y^2$ or $x^2 = (M/y)^2 = 2y^2-1$ . Therefore, we wish to find solutions to the Pell equation $x^2 - 2y^2 = \pm 1,$ which relate to the convergents of the continued fraction representation of $\sqrt{2} = 1 + \frac{1}{2+\frac{1}{2+\frac{1}{2+\cdots}}} .$ These are $\frac{1}{1}, \frac{3}{2}, \frac{7}{5}, \frac{17}{12}, \frac{41}{29}, \ldots,$ corresponding to the solutions $(x,y) = \{(1,1), (3,2), (7,5), (17,12), (41,29), \ldots \}.$ Note that since $N \le 999$ , $2y^2 < 1000$ , or $y < 23$ . Therefore, we have four solutions: $N = \{1, 8, 49, 288\}$ , the sum of which is $\boxed{346}$ .

We have $1+2+\cdots+N = \frac{N(N+1)}{2}$ . Suppose that $\frac{N(N+1)}{2} = k^2 \Leftrightarrow N(N+1) = 2k^2 \Leftrightarrow (2N+1)^2-8k^2=1$ . Let $2N+1 = x$ , we have $x^2-8k^2=1$ . The fundamental of the Pell's equation $x^2-8k^2=1$ is $(x_1,k_1)=(3,1)$ . Therefore, the general solution is given as follow : $\begin{cases} x_{i+1} = 3x_i + 8k_i \\ k_{i+1} = 3k_i + x_i \end{cases}$ Since $1 \le N \le 999$ , we have $3 \le x \le 1999$ . Therefore, we can find the following solutions : $(x_1,k_1) = (3,1), \, (x_2,k_2) = (17,6), \, (x_3,k_3) = (99,35), \, (x_4,k_4) = (577,204).$ The corresponding values of $N$ are : $1,8,49$ and $288$ . The sum of these values are $346$ .

We have: $\displaystyle \sum_{i=1}^N i=\frac{N(N+1)}{2}=s^2$ with s in an integer.

Or $N(N+1)=2s^2$ .

Suppose N is devided by the prime number k, then N+1 is not devided by k, Therefore, each prime number cannot appear in Prime Factorization of both N and N+1. Therefore, in N and N+1, there are a perfect square and a double of perfect square, we have two cases:

Cases 1: $N=2p^2, N+1=q^2$ , so $q^2-2p^2=1$ . According to $[ Pell's Equation ] ( http://mathworld.wolfram.com/PellEquation.html )$ and the condition $1 \leq N \leq 999$ , we find two values of N: 8 and 288.

Cases 2: $N=p^2, N+1=2q^2$ . Similarly, we have two values of N: 1 and 49.

The sum of four values of N: 1+8+49+288=346, so the answer is 346

The explicit formula for the sum $1+2+3+...+N$ is, of course, $\frac{N\cdot(N+1)}{2}$ . In order for this to be a square, either both $N$ and $\frac{N+1}{2}$ are square or both $N+1$ and $\frac{N}{2}$ are square because $N>1\to\gcd(N,N+1)=1\to$ $\gcd(\frac{N}{2},N+1)=1,\gcd(N,\frac{N+1}{2})=1$ .

If $\frac{N+1}{2} = x$ is square, then $2x-1 = N$ must be square. We remind ourselves that $N\leq999 \to x\leq499$ and quickly check to find that only $x=1,25$ satisfy these conditions. Therefore, $N=1,49$ in this case.

If $\frac{N}{2} = y$ is a square, then $2y+1=N+1$ is also a square. Again, $y\leq499$ , and we find that only $y=4,144$ work. Therefore, $N=8,288$ in this case.

All the values of $N$ which satisfy the given conditions are then $1,8,49,288$ and so we report the last three digits of $1+8+49+288=346$ , which are, of course, $\fbox{346}$ .

N and N+1 cannot share a divisor. So sum can be expressed as N (N+1)/2=2a^2 b^2 with N=2a^2 or N=b^2 and gcd (a, b)=1. Now find all perfect squares of the form 2a^2 +/- 1. Checking 0 < a < 23 gives solutions N = 1, 8, 49 or 288.

Every triangular number is of the form $\frac{t(t + 1)}{2}$ . Now, we want to solve

$\frac{t(t+1)}{2} = s^{2}$

With a bit of algebra this becomes $(2t+1)^2=8s^2+1$

and then letting $x$ = 2 $t$ + 1 and $y$ = 2 $s$ , we get

$x^2 - 2y^2 = 1$ .

This is Pell's equation, and the solutions for this are $x = P_{2k} + P_{2k - 1}, y = P_{2k}$ , where $P_{k}$ is the kth Pell Number . Now, the solutions for $N$ are $\frac{1}{2}(P_{2k} + P_{2k - 1} - 1)$ .

The values of $N < 1000$ that satisfies the condition are $1, 8, 49, 288$ , with sum $346$ .

The sum of the series 1+2+3+4+...+N is given by S = N*(N+1)/2

In order for S to be a perfect square we can consider 2 distinct cases,

1) N is a perfect square and (N+1)/2 is a perfect square

1<=N<1000. Also,if (N+1)/2 is to be an integer, N should be an odd number.

Hence the numbers we can consider for N are

1,9,25,49,81,121,169,225,289,361,441,529,625,729,841,961

The ones for which (N+1)/2 is a perfect square are 1,49

2) N+1 is a perfect square and N/2 is a perfect square

N+1 is a perfect square and since N will be divisible by 2,it must be an even number.

So,the values of N to be considered will be

8,24,48,80,120,168,224,288,360,440,528,624,728,840,960

Of which only 8,288 satisfy the condition N/2 is a perfect square

Hence N can take the values 1,8,49,288

So,the sum of all the values of N is = 1+8+49+288=346

The sum N(N+1)/2 is a perfect square. Case 1 : If N is even, then both N/2 and (N+1) should be squares. By trial, we see, N = 8, 288. Case 2 : In N is odd, then both N and (N+1)/2 should be squares. By trial, we see, N = 1, 49. So the answer = 1+8+49+288 = 346

In Python:

import math

def is_square(n):
    return math.sqrt(n).is_integer()

if __name__ == '__main__':
    squares = []
    for n in range(999):
        sum = 0
        for n2 in range(n+1):
            sum += n2+1
        if is_square(sum): squares.append(n+1)
    allsum = 0
    for n in squares:
        allsum += n
    print(str(allsum)[-3:])

N(N+1)/2=X^2 with X in an integer.(given condition) Or N(N+1)=2X^2 as we know if N is divided by the prime any number k, then N+1 is not divided by k, Therefore, each prime number cannot appear in Prime Factorization of both N and N+1. so, N and N+1, there are a perfect square and a double of perfect square now we can have two conditiions N=2p^2 and N+1=q^2 or q^2−2p^2=1. its a PELL'S equation Refer en.wikipedia.org/wiki/Pell...'s equation and en.wikipedia.org/wiki/Pell... number

and the condition 1≤N≤999, two values of N are 8 and 288. Cases 2: N=p^2,N+1=2q^2. again weget two values of nN: 1 and 49.so sum of 4 values of N is 1+8+49+288=346, so the answer is 346

N(N+1)/2=X^2 with X in an integer.(given condition) Or N(N+1)=2X^2 as we know if N is divided by the prime any number k, then N+1 is not divided by k, Therefore, each prime number cannot appear in Prime Factorization of both N and N+1. so, N and N+1, there are a perfect square and a double of perfect square now we can have two conditiions N=2p^2 and N+1=q^2 or q^2−2p^2=1. its a PELL'S equation Refer en.wikipedia.org/wiki/Pell...'s equation and en.wikipedia.org/wiki/Pell... number

and the condition 1≤N≤999, two values of N are 8 and 288. Cases 2: N=p^2,N+1=2q^2. again weget two values of nN: 1 and 49.so sum of 4 values of N is 1+8+49+288=346, so the answer is 346