4
$\begingroup$

Can someone help me with this exercise? I tried to do it, but it was very hard to solve it.

Find the value of $n$ to make $2^8 + 2^{11} + 2^n$ a perfect square.

It is the same thing like $4=2^2$.

$\endgroup$
1
  • $\begingroup$ $n = 12$ gives $6400 = 40^2$ $\endgroup$ – SS_C4 Feb 20 '16 at 12:24
6
$\begingroup$

$2^8 + 2^{11} + 2^n = 2^8(1 + 8 + 2^{n-8})=2^8(9 + 2^{n-8})$

Therefore, $9 + 2^{n-8}$ has to be a perfect square. Clearly, $9 + 16 = 25$ is a perfect square.

So, $2^{n-8} = 2^4$ giving, $$n = 12$$

$\endgroup$
6
  • $\begingroup$ Best answer thankful $\endgroup$ – Mang Sopheak Feb 20 '16 at 12:45
  • 2
    $\begingroup$ @MangSopheak It doesn't prove that $n=12$ is the only solution. $\endgroup$ – user236182 Feb 20 '16 at 12:47
  • $\begingroup$ So can u prove it? $\endgroup$ – Mang Sopheak Feb 20 '16 at 12:48
  • $\begingroup$ @MangSopheak My answer proves it. Also MXYMXY's. $\endgroup$ – user236182 Feb 20 '16 at 12:49
  • $\begingroup$ Yeah u are right the n has more answer $\endgroup$ – Mang Sopheak Feb 20 '16 at 13:01
21
$\begingroup$

Hint: $(2^a + 2^b)^2 = 2^{2a} + 2^{2b} + 2^{a+b+1}$.

$\endgroup$
2
  • 2
    $\begingroup$ This only finds one possible solution but doesn't prove there are no others. $\endgroup$ – user236182 Feb 20 '16 at 12:36
  • 2
    $\begingroup$ @user236182 he did not ask for that. I just wanted to show the idea that taking the square of a sum of 2-powers has the correct form. $\endgroup$ – Henno Brandsma Feb 20 '16 at 12:37
10
$\begingroup$

If $0\le n\le 7$, then there are no solutions. Let $n\ge 8$.

$$2^8+2^{11}+2^n=\left(2^4\right)^2\left(9+2^{n-8}\right)$$

is a square if and only if $9+2^{n-8}=m^2$ for some $m>3$, i.e. $2^{n-8}=(m+3)(m-3)$, so $m+3=2^k$ and $m-3=2^l$ for some $k>l\ge 0$. If $k\ge 4$, then $$6=2^k-2^l\ge 2^k-2^{k-1}\ge 8>6$$ contradiction, so $k\in\{1,2,3\}$, which only gives $k=3$, so $m=5$, $n=12$.

$\endgroup$
4
  • 1
    $\begingroup$ How do I know without calculating that for $0<=n<=7$ there are no solutions? Or the only way is calculating? $\endgroup$ – Shubhraneel Pal Jul 23 '17 at 13:43
  • 2
    $\begingroup$ @ShubhraneelPal If $0\le n\le 7$, then $2^8+2^{11}+2^n$ $=2^n(2^{8-n}+2^{11-n}+1)$, where $11-n>8-n>0$, $2^{8-n}+2^{11-n}+1$ is odd, so if $2^n(2^{8-n}+2^{11-n}+1)$ is a perfect square, then $n$ is even. So you only need to check $n\in\mathbb \{0,2,4,6\}$. If $n\in\{2,6\}$, then $2^{8-n}+2^{11-n}+1\equiv 2\pmod{5}$, $2$ is not a quadratic residue mod $5$. If $n=4$, then $2^{8-n}+2^{11-n}+1\equiv 5\pmod{7}$, $5$ is not a quadratic residue mod $7$. If $n=0$, then $2^{8-n}+2^{11-n}+1\equiv 6\pmod{11}$, $6$ is not a quadratic residue mod $11$. $\endgroup$ – user236182 Jul 24 '17 at 15:14
  • 2
    $\begingroup$ @ShubhraneelPal For $n\in\{0,4\}$ you could also notice that $\frac{2^{8-n}+2^{11-n}+1}{5}$ is not divisible by $5$ (i.e. doesn't end with the digit $0$ or $5$). There are various methods. $\endgroup$ – user236182 Jul 24 '17 at 16:29
  • 2
    $\begingroup$ @ShubhraneelPal If $n\in\{2,6\}$, then $2^{8-n}+2^{11-n}+1$ is a prime number, so not a perfect square. $\endgroup$ – user236182 Jul 24 '17 at 16:45
4
$\begingroup$

Note the fact that $2^8+2^{11}=48^2$

This implies that we are trying to find values of $n$ where $2^n=(x-48)(x+48)$.

Thus, we must find $k,l$ where $2^k-2^l=96$(where $x+48=2^k$, $x-48=2^l$)

Note the fact that $k \ge 7$.

This implies that $2^k$ is divisible by $32$, which implies that $2^l$ is also divisible by $32$.

Also, notice that if $k \ge 9$, when $2^k-2^l \ge 256$.

This implies that $k=7$or $k=8$.

Note the fact that $k=8$ does not have an integer solution, thus $k=7$.

Thus $x=80$, and thus $n=12$.

$\endgroup$
4
$\begingroup$

See we need an integer power of $2$ so difference between a square and know indices should be even now $2$ follows a particular order of last digits which is $2,4,8,6...$ so now we need numbers which are even which will give difference as these above numbwrs .Now we know no perfect square ends in $2,8$ so now we are left with $4,6$ but(last digit) $4-4=0$ no power of $2$ gives $0$ as last digit so now we need to check for $6$ now when we plug $60^2,70^2..$ we get our last digit as $6$ ie eg $3600-2304=...6$ so now checking these cases we get $n=80$ so $80.80=6400$ $2^8+2^{11}=2304$ so $6400-2304=4096$ which is $2^{12}$ thus $n=12$

$\endgroup$
1
$\begingroup$

You can write

$$2^8+2^{11}+2^n=(2^4)^2+2.2^4.2^6+2^n$$

now, note that if $n=12$, follows that

$$(2^4)^2+2.2^4.2^6+2^{12}=(2^4)^2+2.2^4.2^6+(2^6)^2=(2^4+2^6)^2$$

Thus, $n=12$ to solve the problem.

$\endgroup$
1
$\begingroup$

Here is a very late answer since I just saw the problem:

By brute force, we may check that $n=12$ is the smallest possible integer such that $2^8+2^{11}+2^n$ is a perfect square. We also claim that this is the only integer.

To see why the above is true, let $2^8+2^{11}+2^n=2^8(1+2^3+2^k)=2^8(9+2^k), k \ge 4$. Now, we only need to find all integers $k \ge 4 $ such that $9+2^k$ is a perfect square. Let $m^2=9+2^k, m \ \in \mathbb{Z^+}$. Hence we have $2^k=(m-3)(m+3)$. Clearly, $m$ must be odd; otherwise, both $m-3$ and $m+3$ would be odd, contradiction. Hence, $2\mid m-3$ and $2 \mid m+3$. But we also claim it is impossible that both $m-3$ and $m+3$ divides $4$; otherwise, $4 \mid 2m \Rightarrow 2 \mid m$, which is a contradiction for $m$ odd. Since $m-3 < m+3$, we conclude that, for any integer $k \ge 4$, we must have $m-3=2$ and $m+3=2^{k-1}$. But the former already implies $m=5$, so we must have $k-1=3 \Rightarrow k=4$ as our only solution, which is precisely what we claimed earlier.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.