International olympiad of metropolises 2016 P1

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Source: International olympiad of metropolises 2016

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#1 h Sep 7, 2016, 8:59 PM • 2 Y

Y by Adventure10, Mango247

Find all positive integers $n$ such that there exist $n$ consecutive positive integers whose sum is a perfect square.

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#2 h Sep 8, 2016, 7:16 AM • 2 Y

Y by Adventure10, Mango247

say the starting integer is a, then the sum is $\frac{n(n+2a-1)}{2}$
note that if n is odd and $a=\frac{n+1}{2}$ , the sum is then $n^2$ , so there are infinitely many n, not sure when n is even, still calculating

This post has been edited 1 time. Last edited by cyborg_w, Sep 8, 2016, 7:17 AM

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#3 h Sep 8, 2016, 11:59 AM • 1 Y

Y by Adventure10

Write $n$ in the form $n=2^k(2m+1)$ with integers $k,m\ge0$ .
Answer: such a square exists if and only if (i) $k=0$ or (ii) $k$ is odd.

(1) If $k=0$ , then pick $a=m+1$ .

(2) If $k$ is odd, then let $x^2$ be a square that is strictly larger than $2^k$ .
Pick $a$ so that $2a-1=(2m+1)(y^2-2^k)$ .

(3) If $k\ge2$ is even, then $\frac12n(n+2a-1)$ is divisible by $2^{k-1}$ but not $2^k$ , and cannot be a square.

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#4 h Sep 13, 2016, 7:38 PM • 4 Y

Y by Pluto1708, HansV, Adventure10, Mango247

Notice that the sum of all these numbers, say $k+1,\dots, k+n$ equals $S=nk+\frac{n(n+1)}{2}$ . Now, if $n$ is odd we set $n=2m-1$ . Clearly, setting $k=(2m-1)\cdot s^2-m$ for a sufficiently large $s$ yields that $S=((2m-1)s)^2$ is a perfect square. If $u=v_2(n)>0$ then set $n=2^u\cdot v$ for $v$ odd and see that $S=2^{u-1}\cdot (2k+v\cdot (2^u\cdot v+1))$ and $v_2(S)=u-1$ must be even. Thus $u$ must be odd. Now, take $k=\frac{(2l+1)^2-v\cdot (2^u\cdot v+1)-1}{2}$ for $l$ sufficiently large. Clearly, $k$ is an integer and $S=(2^{\frac{u-1}{2}}\cdot l)^2$ is a perfect square. The discussion concludes and we see that the only such $n$ are those for which $v_2(n)$ is either $0$ or an odd number.

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#5 h Jul 12, 2024, 2:29 PM

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The answer is when $\nu_2(n)$ is $0$ or odd. Firstly, if $\nu_2(n) = 0$ , then if we let $n = 2k + 1$ , we have $(k + 1) + (k + 2) + \cdots + (3k + 1) = (2k + 1) ^2$
Now assume $\nu_2(n) > 0$ . Let $n = 2k$ for some positive integer $k$ . Any $n$ consecutive integers must be of the form $a, a+1, \ldots, a + (2k - 1)$ , Now, $a + (a + 1) + \cdots + (a + 2k - 1) = (2a + 2k - 1) \cdot k$ Notice that if $\nu_2(n)$ is even, then $\nu_2(k)$ is odd, and $\nu_2(2a + 2k - 1) = 0$ , so $\nu_2((2a + 2k - 1) k )$ is odd, and thus $(2a + 2k - 1) \cdot k$ can't be a perfect square. Thus, $\nu_2(n)$ is odd if $\nu_2(n) > 0$ . It suffices to show that all such $n$ work.

Let $k = 2^{2c} \cdot r$ , where $r$ is odd. Now let $N^2 > 2k - 1$ be a sufficiently large odd perfect square. We may choose $2a + 2k - 1 = N^2 \cdot r$ , since $2a + 2k - 1$ can take any odd number greater than $2k - 1$ . Now, we have $(2a + 2k - 1) \cdot k = 2^{2c} \cdot N ^2 \cdot r^2$ , which is a perfect square, as desired.

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