How do I solve Problem 1 of the International Olympiad of Metropolis? The question is: Find all $n$ so that there exists $n$ consecutive numbers whose sum is a square!
My method to solve the problem: I would try to look at the values modulo $n$, and there I see there is no possible square for $n=2x>2$, so all even numbers bigger than 2, and that all odd numbers work...
I can also see that the sum of the consecutive numbers is $$\dfrac{((x+1)+(x+n))
\cdot n}{2}$$ So that has to give a square...
But how do I do that? I cannot put my method in application... Am I totally wrong, and I shouldn't do it like that?
 A: As you noted in the question, if $x +1$ is the first number in the sum, then the sum of the $n$ consecutive numbers is
$$ \frac{n}{2} \cdot (x + 1 + x + n). $$
If $n$ is odd, let $n = 2k + 1$, and take $x = k$. Then the sum of the $n$ consecutive numbers $(k + 1), (k + 2), \dots, (k + n)$ is
$$
  \frac{2k+1}{2} \cdot (k + 1 + k + 2k + 1) = (2k + 1)^2
$$
which is a square, so all odd numbers work.
Now suppose that $n$ is even. Let $n = 2k$ for some $k$.
Then the sum is equal to
$$ k(2x + 2k + 1) $$.
The term in brackets is odd, so for this to be a square, the largest power of $2$ which divides $k$ must be even. i.e. We must have that $k = 2^{2m} \cdot d$ where $m$ is some natural number, and $d$ is an odd natural number.
The sum then reduces to
$$ 2^{2m} \cdot d \cdot (2x + 2^{2m + 1} \cdot d + 1) $$
This is a square if and only if
$$ d \cdot (2x + 2^{2m + 1} \cdot d + 1) $$
is a square, so it is enough to choose $x$ so that the term in brackets is a square multiple of $d$. That is, we try to choose $x$ so that
$$ 2x + 2^{2m + 1} \cdot d + 1 = a^2 \cdot d $$
for some natural number $a$.
This is equivalent to
$$ 2x + 1 = \left( a^2 - 2^{2m + 1} \right) \cdot d $$
and it is easy to see that this has a solution for x as long as $a$ is an odd square greater than $2^{2m + 1}$. Thus we have a solution in this case since there are arbitrarily large odd square numbers.
We see that if $n$ is even, then $n$ works if and only if $n$ is divisible by $2$ an odd number of times.
A: As noted in the statement of the question, the sum of $n$ consecutive numbers starting from $x+1$ is given by
$$\sum_{j=x+1}^{x+n}j=\frac{(2x+1+n)n}{2}.$$
If $n=2m+1$ is an odd integer, then we require
$$(x+m+1)(2m+1) = k^2$$
for some integer $k$. Clearly taking $x=m$ will yield a solution (as also noted by Win Vineeth). In fact, there are infinitely many solutions simply by solving for $x$, giving
$$x = \frac{k^2}{n} - (m+1).$$
Then every $k$ such that $n\mid k^2$ will yield a distinct solution $x$, and conversely, these yield all the valid solutions.
On the other hand, if $n=2m$ is an even integer, then we require
$$(2x+2m+1)m=k^2.$$
Let $2^s$ be the largest power of $2$ which divides $m$. Then writing $m=2^sm'$ where $m'$ is odd, we have
$$(2x+2^{s+1}m'+1)2^sm'=k^2.$$
It follows that $s$ must be even, so that if $n$ is even, it must contain an odd number of $2$s in its factorization. Writing $k^2 = 2^sk'^2$ where $k'$ is odd, we then get
$$(2x+2^{s+1}m'+1)m'=k'^2.$$
Then again there are an infinite family of solutions
$$x=\frac{k'^2-m'}{2}-2^{s}m',$$
where every odd $k'$ will yield a distinct solution. Again, these yield all valid solutions.
In summary, there is a solution for $n$ if and only if $n$ is odd or if $n$ is even with an odd number of $2$s in its prime factorization. When there is a solution, there are infinitely many solutions.
A: All numbers other than odd multiples of $2^m$, where m is even.
Sum of n terms will be $ n/2 * ((2*x)+(n-1)) $
When m is even, the denominator is cancelled and 2 will be again raised to odd power in numerator and term in brackets will be odd number.
So you cannot have such n. 
A: For any odd $n$, sum of $n$ consecutive numbers starting from $\frac {n+1}2$ is 
$$\frac n2 (n+1+(n-1)*1) = n^2$$
Thus for every odd $n$, there exist $n$ consecutive numbers whose sum is a perfect square.
More details - This becomes simpler if you know Arithematic Progression. The formula for sum of $n$ numbers starting from $a$, which are $d$ apart (difference between successive numbers is $d$) is 
$$S_n=\frac n2(2a+(n-1)d)$$
Here $d=1$ and $a$ can be calculated to be $\frac {n+1}2$
Thank you @Litho for pointing it out.
