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.
 A: I don't know what tools you are supposed to use to demonstrate that 4 is not a square-positive number. I guess you can do it in different ways. Here's one:
Take four consecutive numbers.
Now take a unit from the last one and put it to the second.
The last three numbers are now equal. If you sum them all, you get: $3n + (n - 2) = 4n - 2$.
Now for reduction ad absurdum you want that expression to give you the square of an integer: $4n - 2 = x^2$ which is equivalent to: $n = ( x^2 + 2) / 4$. Now you can say that n is an integer only if $( x^2 + 2 )$ is multiple of 4, but this can be true only if x is an even number. Now the square of any even number is of the form $(2^2n *...)$ i.e. $(4^n *...)$ So it is always a multiple of 4. If $x^2$ needs to be always divisible for 4, then $(x^2 + 2)$ can never be a multiple of 4 and therefore n can never be an integer. But $n$ is of course an integer, it is the third of the four consecutive numbers chosen.
A: There is a pretty simple test to show that $n=4q$, where q is odd, cannot be a solution.
The sum of the first $a$ integers is:
$$\frac{a(a+1)}{2}$$
so the sum of $n$ consecutive integers starting at $a$ is:
$$\frac{(a+n)(a+n-1)}{2}-\frac{a(a-1)}{2} = \frac{n(2a+n-1)}{2}$$
since this number must be equal to a perfect square:
$$\frac{n(2a+n-1)}{2}=k^2$$
substitute $n=4q$ for some odd integer $q$:
$$2q(2a-1+4q)=k^2$$
it should be apparent that $k$ must be an even integer, therefore substitute $k=2m$:
$$2(2a-1+4q)q=4m^2$$
$$(2a-1+4q)q=2m^2$$
The numbers $2a-1+4q$ and $q$ must be odd, but the RHS must be even, therefore it is not possible for this equation to be solved over the integers, and so there is no solution for $n=4q$.
I know that this isn't a full proof, but I am still working on it
A: We prove the following result:
If $n=2^b \cdot d, d \text{ is odd, then } n \text{ is square-positive if and only if } b=0  \text{ or } b  \text{ is odd}$.
$\underline{\text{Case 1}}$: $b$ is even and $b>0$. Let $b=2c$. We prove by contradiction $n$ cannot be square-positive. If it were, then there exist $n$ consecutive positive integers,
$$
a, a+1, \ldots, a+n-1
$$
such that the sum of these numbers is a square, i.e.,
$$
S=a+(a+1)+\cdots + (a+n-1) = n\frac{n+2a-1}{2} =T^2
$$
for some $T\in \mathbb{N}$.
Note that $n=2^{2c}\cdot d$ is even, and
$$
T^2 = S = 2^{2c-1}\cdot d \cdot (n+2a-1),
$$
which cannot be a perfect square because 1) $d(n+2a-1)$ is odd and, 2) $2c-1$, the highest power of its factor $2$ is odd, a contradiction.
$\underline{\text{Case 2}}: b=0$. Let $n=d=2m+1$, then the sum of the following $n$ consecutive numbers is a perfect square:
$$
n-m, n-m+1, \ldots, n+m.
$$
Indeed,
$$
S = (n-m)+(n-m+1)+\cdots +(n+m) = n \cdot \frac{(n-m)+(n+m)}{2}=n^2.
$$
$\underline{\text{Case 3}}: b=2c+1 $ for some $c\geqslant 0$, then $n=2^{2c+1}d$. We show that the sum of the following $n$ consecutive numbers is a perfect square:
$$
a, a+1, \ldots, a+(n-1),
$$
where
$$
a=\frac{1+d(K^2-2^{2c+1})}{2}
$$
and $K$ is some odd positive integer such that
$$
K^2 > 2^{2c+1}.
$$
For example, if $c\geqslant 1$, we can pick $K=2^{c+1}-1$. If $c=0$, we can pick $K=3$. Also note that $a$, the first number in the sequence, is a positive integer by construction.
Finally,
$$
S=n\frac{n+2a-1}{2} = \frac{2^{2c+1}d}{2} \left[ 2^{2c+1}d + d(K^2-2^{2c+1}) \right] = 2^{2c} d [ d K^2 ] = (2^c d K)^2. \blacksquare
$$
Example: $n=8=2^3 \cdot 1, c=1, d=1, K=3 \Rightarrow a=1$, and the sum of the sequence is $1+2+\cdots+8=36=6^2$.
A: 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.
=====
In general:
........
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.
