IstekOlympiadTeam wrote:

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The number is odd.

Prove that $ n=3 $.

IstekOlympiadTeam wrote: Submathematics wrote: IstekOlympiadTeam wrote:

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The number is odd. NOT!!! So, you are still saying $1^3+2^3+...16^3+17^n$ is even? Sir, are you here to troll?

Hi! You get $ [ \frac {16(16+1)}{2} ]^2+ 17^n= k^2 $ so $ (k- [ \frac {16(16+1)}{2} ])(k+ [ \frac {16(16+1)}{2} ])=17^n $ and from here it is obvious.

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sol

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JBMO Shortlist 2013 N1 wrote: $\boxed{N1}$ find all positive integers $n$ for which $1^3+2^3+\cdots+{16}^3+{17}^n$ is a perfect square. Solution:- Assume that $1^3+2^3+\cdots+16^3+17^n=k^2\implies 17^2(17^{n-2}+8^2)=k^2$. Note that $\gcd(17^2,(17^{n-2}+8^2))=1$, hence, $17^{n-2}+8^2$ must also be a perfect square. Let $n-2=m$ for our sake of convenience. So, now the problem reduces to find:- All $m$ where $m\in\mathbb Z^+$ such that $17^m+8^2=k^2$. So, $17^m=(k+8)(k-8)$, now notice that the difference between $(k+8),(k-8)$ is $16$. Hence, if $17|(k+8)$ then $17\nmid (k-8)$ and viceversa. So, if $17|k+8\implies k+8=17x$ for some $x\in\mathbb Z^+$. So, putting this in the equation we get that $17^{m-1}=x(k-8)$, now notice that $17$ is a prime number, hence prime decomposition of $17^{m-1}$ will not have any prime factors other that $17$, but $17\nmid{k-8}$, so this forces $k-8=1\implies k=9\implies m=1\implies n=3$. Hence, $n=3$ is the only solution.

Note that $1^3 + 2^3 + ... + 16^3 = \left(\frac{16 \cdot 17}{2} \right)^2 = 8^2 \cdot 17^2$. Therefore, we want $8^2 \cdot 17^2 + 17^n$ to be a perfect square. It is easy to show that $n = 1,2$ doesn’t satisfy this constraint, so from here on, let’s assume $n \ge 3$. Then, we want $8^2 + 17^{n-2}$ to be a perfect square. Mathematically, we can write this as: $$8^2 + 17^{n-2} = x^2 \Longleftrightarrow (x-8)(x+8)=17^{n-2}$$for some $x \in \mathbb{Z}$. If $x - 8 = 1$, then $x = 9$, which does indeed produce the solution $n =3$. If $x + 8= 1$, then no solution is produced. Therefore, for all $n \ge 4$, $x-8 \ne 1$ and $x + 8 \ne 1$. It follows that $x - 8$ and $x+8$ must both be divisible by $17$, but this is absurd given that one implies $x \equiv 8 \mod 17$ and the other implies that $x \equiv 9 \mod 17$. It is proved that the only solution is $\boxed{x=3}$.