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Find $n \in \mathbb{N}$ such that $\dfrac{n^3+7}{2^n}$ is perfect square.

We need $n^3+7 \equiv 0 \pmod{2^n}$ and also make sure that the fraction is a perfect square. How do we find such $n$?

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  • $\begingroup$ you want it to be an integer? $\endgroup$ – Jorge Fernández Hidalgo Dec 25 '16 at 1:30
  • $\begingroup$ @JorgeFernándezHidalgo If the fraction is a perfect square it needs to also be an integer. $\endgroup$ – user19405892 Dec 25 '16 at 1:31
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    $\begingroup$ Trying $n = 1$ gives $4$, which is a perfect square. $\endgroup$ – Mark Dec 25 '16 at 1:31
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Hint Prove that if $n \geq 10$ then $$2^n > n^3+7$$

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For integers $n \ge 10$, you can prove that $0 < n^3+7 < 2^n$, and so, $0 < \dfrac{n^3+7}{2^n} < 1$. Thus, $\dfrac{n^3+7}{2^n}$ can't be a perfect square.

For integers $n \le -2$, we have $n^3+7 < 0 < 2^n$, and so, $\dfrac{n^3+7}{2^n} < 0$. Thus, $\dfrac{n^3+7}{2^n}$ can't be a perfect square.

Testing $n = -1,0,\ldots,8,9$, we see that $\dfrac{n^3+7}{2^n}$ is only a perfect square for $n = 1$.

EDIT: It appears that the title says "$n \in \mathbb{Z}$" while the question says "$n \in \mathbb{N}$". If the question was is indeed $n \in \mathbb{N}$, then we can ignore the case where $n \le -2$.

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Notice that $11^3+7<2^{11}$. It is easy to prove with induction that $11^N+7<2^{N}$ for $N\geq 11$.

Hence we mus try only with numbers in the range $0,1,2,3,4,5,6,7,8,9$.

Of course all even values except $0$ need not be considered, as the numerator would be odd, and $0$ does not work.

So we are left only with $1,3,5,7,9$.

$n=1$ works.

$n=3$ does not work as $34$ is not a multiple of $8$.

$n=5$ does not work as $126$ is not a multiple of $32$

$n=7$ does not work as $344$ is not a multiple of $128$

$n=9$ does not work as $730$ is not a multiple of $512$

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Let $\displaystyle f(n)={n^3+7\over 2^n}$. Then we have $$f(0)= 7$$ $$f(1)= 4$$ $$f(2)= 15/4$$ $$f(3)= 17/4$$ $$f(4)= 71/16$$ $$f(5)= 33/8$$ $$f(6)= 223/64$$ $$f(7)= 175/64$$ $$f(8)= 519/256$$ $$f(9)= 23/16$$ $$f(10)= 1007/1024.$$ Of these values, the only perfect square is $f(1)= 4$. Also, we easily see that $0<f(n)<1$ when $n\ge10$ because the denominator $2^n$ grows faster than the numerator $n^3+7$ (and eventually $2^n$ grows faster than any polynomial function of $n$). Thus the only $n$ satisfying the given conditions is $n=1$.

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