An integer $n \geq 2$ is called square-positive- proof? An integer $n \geq 2$ is called square-positive if there are $n$ consecutive positive integers whose sum is a square. Determine the first four square-positive integers.
So I have found the first four square-positive numbers, but I need to prove that why it $4$ is not a square-positive number and I also need to write a general formula for determining whether a number is square-positive or not. I have tried to write the sum of consecutive positive integers like this $a + a +1 + a + 2 + a+3 \dots a - 1$ and I wrote it like this for all numbers, and part of the proof for why $4$ isn't a square-positive number is that $4a + 6$ is not divisible with $4$. But I haven't got so far.
Here is my answer:
2 : 4 + 5 = 9  which is 3^2
3 : 2 + 3 + 4 = 9 which is 3^2
5: 18 + 19 + 20 + 21 + 22 = 100  which is 10^2
6: 35 + 36 + 37 + 38 + 39 + 40 = 225  which is 15^2
Interesting fact is that for all odd numbers and some even numbers like 6 and 10, you can find out which number is the first (the one you start with and then go forward here like 3, 2, 18 and 35) using this formula :
(I show it in an example because I still can't write it algebraically):
For example: the sum of 95 subsequent numbers is 5n + 10
(10^2 - 10) /5 = 18
So your first number is 18
And if you keep adding, 18 + 19 + 20 + 21 + 22 you get 100 which is 10^2, the same number you squared.

Let $b-2, b-1, b, b+1$ be four consecutive numbers. The sum is $4b -2$.  Let $m^2 = 4b-2$  Then $\frac {m^2}2 = 2b-1$.  And $2|m^2$.  So $2|m$.  Let $m=2k$ then $4k^2 = 4b -1$ and $2k^2 = 2b-1$.  That's impossible as the LHS is even and the RHS is odd.  So $4$ is not square positive.
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In general:
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If the first integer is $a$ and there are $n$ consecutive integers the numbers are $a, a+ 1,......, a+(n-1)$ and the sum of the the integers is $a + (a+ 1) + ..... + (a + (n-1)) = n\cdot a + (1+ 2 + ...... + (n-1)) = n\cdot a + \frac {(n-1)n}2$.
If $n\cdot a + \frac {(n-1)n}2 = M^2$ then by completing the square
$\frac 12n^2 - (a-\frac 12)n -M^2 = 0$ and .....
well, we can keep going but.... let's try to be clever....
If $n=2k+1$ is an odd number and the middle number is $b$ then the first number is $b-k$ and the last number is $b+k$ and the sum of the number is $$(b-k) + (b-k+1) + ..... + (b-1) + b+ (b+1) + ..... + (b+k) =\\ nb + (-k+(-k+1)+...... + (-1) + 0 + 1 + 2 + ..... k) =\\nb + 0= nb$$.  ANd $nb$ can be square positive whenever $b= n$.  (and other times).  For intsance.  $1=1^2$ and $2+3+4=3^2$ and $3+4+5+6+7 = 5^2$ and $(n-\frac {n-1}2) + ...... + (n + \frac {n-1}2) = n^2$.
So all odd numbers are square positive.
And if $n= 2k$ is even the there is no middle number but if the "middlish" number (off by one half) is $b$:Then we can notice that first $k$ of these numbers are $b-k ,b-(k-1) , b-(k-2),........, b-1$; and the second $k$ of these numbers are $b, b+1, b+2, ......, b+(k-1)$; and the sum of these numbers are $$nb + (-k,-(k-1),-(k-2),......., (k-2),(k-1)) =\\ nb - k =\\ 2kb -k = k(2b-1)$$.  And that can be a square positive every time we have $2b-1 = k=\frac n2$.
But this is possible only if  $\frac n2$ is odd.
But if $k=2m$ is even we must have $2m(2b-1)$ is a perfect square which requires $m$ to be even.
By induction will prove $n$ is square positive if and only if $n = 2^{odd}\cdot odd$.
So the list of all square positive numbers are $2,3,5,6,7,8,9,10,11,13,14,15,17,18,19,21,22,32,24,......$ etc.
