WakeUp wrote: Find the positive integers $n$ that are not divisible by $3$ if the number $2^{n^2-10}+2133$ is a perfect cube.

hint: being n not divisible by $3$ ==> $n^2=3k+1$ .......(*) equation can get form: $x^3-(2^{k-3})^3=2133$, where $2133=3^3*79$ $(x-3)(x^2+x*2^{k-3}+(2^{k-3})^2)=3^3*79$ etc...

roza2010 wrote: WakeUp wrote: Find the positive integers $n$ that are not divisible by $3$ if the number $2^{n^2-10}+2133$ is a perfect cube.

hint: being n not divisible by $3$ ==> $n^2=3k+1$ .......(*) equation can get form: $x^3-(2^{k-3})^3=2133$, where $2133=3^3*79$ $(x-3)(x^2+x*2^{k-3}+(2^{k-3})^2)=3^3*79$ etc... Litle mistake:$x^3-(2^{k-3})^3=(x-2^{k-3})(x^2+x*2^{k-3}+(2^{k-3})^2)$ We know $n\ge\sqrt{10}$ and when $n\ge5$, $k\ge8$ and so $(x-2^{k-3})(x^2+x*2^{k-3}+(2^{k-3})^2)\ge3*2^13 \ge (33^2+33*32+32^2)>3000>2133$ So the only one is $4$, we show that this is correct: $2197=13^3$

the condition $3|n$ is waste. for if $3|n$,then $2^{n^2-10}+2133\ne 1 or -1(mod 9)$ so it can't be a cube

I think this solution si correct as well: We try all possible $n$ from $1$ to $6$ and see that $n=4$ works(*). From now on, we assume $n \ge 5$. We know that $2^{n^2-10}+2133 = a^3$. Note that $n^2 \mod 3 \equiv 1\implies n^2 = 3k + 10$. Thus,$$2^{3k}+2133 = a^3 \implies \left(2^k \right)^3+2133 = a^3 \implies \left( a-2^{k}\right) \left(a^2 + 2^ka + 2^{2k} \right) = 2133.$$Noting that by $n^2=3k+1$, $n \ge 7$, and thus $k \ge 13$, $a \ge 14$, can find lower bounds on $a^2 + 2^k \cdot a + 2^{2k}$: $$a^2 + 2^k \cdot a + 2^{2k} \ge 14^2 + 2^7 \cdot 14 + 2^{14} \ge 2133.$$Hence, the only $n$ that satisfies the given diophantine equation is $n=\boxed{4}$. (*) Note that we don't need to test $n = 3, 6$. When $n=5$, the numbers can get a bit messy if you straight forward bash, but we can prove that $a = 32$ is too low and $a=33$ is too high instead.

AopsUser101 wrote: I think this solution si correct as well: We try all possible $n$ from $1$ to $6$ and see that $n=4$ works(*). From now on, we assume $n \ge 5$. We know that $2^{n^2-10}+2133 = a^3$. Note that $n^2 \mod 3 \equiv 1\implies n^2 = 3k + 10$. Thus,$$2^{3k}+2133 = a^3 \implies \left(2^k \right)^3+2133 = a^3 \implies \left( a-2^{k}\right) \left(a^2 + 2^ka + 2^{2k} \right) = 2133.$$Noting that by $n^2=3k+1$, $n \ge 7$, and thus $k \ge 13$, $a \ge 14$, can find lower bounds on $a^2 + 2^k \cdot a + 2^{2k}$: $$a^2 + 2^k \cdot a + 2^{2k} \ge 14^2 + 2^7 \cdot 14 + 2^{14} \ge 2133.$$Hence, the only $n$ that satisfies the given diophantine equation is $n=\boxed{4}$. (*) Note that we don't need to test $n = 3, 6$. When $n=5$, the numbers can get a bit messy if you straight forward bash, but we can prove that $a = 32$ is too low and $a=33$ is too high instead. Why do you change the values as $n^2=3k+10$ and while $n^2=3k+1$? Then $n\neq n$?

@above, where did I say n^2=3k+1? its possible i said that somewhere because on my paper that's what I write but when typing I was like "why not let n^2 = 3k+10"? plz show me the exact line where error occurs .

AopsUser101 wrote: I think this solution si correct as well: Noting that by $n^2=3k+1$, $n \ge 7$, and thus $k \ge 13$, @above Your solution is correct, but this typo confused me a few, but when you do $3*13+1=40$ you know that you're talking about $3*13+10=49$.

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