# Make $2^8 + 2^{11} + 2^n$ a perfect square

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 $4=2^2$.

• $n = 12$ gives $6400 = 40^2$ Feb 20, 2016 at 12:24

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

• This only finds one possible solution but doesn't prove there are no others. Feb 20, 2016 at 12:36
• @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. Feb 20, 2016 at 12:37

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 $m=5$, $n=12$.

• How do I know without calculating that for $0<=n<=7$ there are no solutions? Or the only way is calculating? Jul 23, 2017 at 13:43
• @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 $11$. Jul 24, 2017 at 15:14
• @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. Jul 24, 2017 at 16:29
• @ShubhraneelPal If $n\in\{2,6\}$, then $2^{8-n}+2^{11-n}+1$ is a prime number, so not a perfect square. Jul 24, 2017 at 16:45

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

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, $$n = 12$$

• Best answer thankful Feb 20, 2016 at 12:45
• @MangSopheak It doesn't prove that $n=12$ is the only solution. Feb 20, 2016 at 12:47
• So can u prove it? Feb 20, 2016 at 12:48
• @MangSopheak My answer proves it. Also MXYMXY's. Feb 20, 2016 at 12:49
• Yeah u are right the n has more answer Feb 20, 2016 at 13:01

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 $n=12$.

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 $n=12$

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.

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.

Solution : $$2^n+2^8+2^(11)=m^2$$ $$2^n=m^2-2^8-2^(11)$$ $$2^n=m^2-2^8(1+2^3)$$ $$2^n=m^2-2^8*9$$ $$2^n=m^2-(2^4*3)^2$$ $$2^n=m^2-48^2$$ $$2^n=(m-48)(m+48)$$

$$2^(n-k+k)= (m-48)(m+48)$$  $$2^(n-k)*2^k= (m-48)(m+48)$$ So (1) $$2^(n-k)= (m-48)$$ and (2) $$2^k= (m+48)$$

(2)-(1)  $$2^k-2^(n-k)=(m+48)-(m-48)$$              $$2^k-2^(n-k)=96$$              $$2^k-2^(n-k)=2^5*3$$              $$2^(n-k)*(2^(2k-n)-1)= 2^5*3$$               $$2^(n-k)=2^5$$  and  $$2^(2k-n)-1=3$$               $$n-k=5$$            and   $$2^(2k-n)=2^2$$               $$n-k=5$$            and   $$2k-n=2$$               $${n=12,k=7}$$

$$n=12$$