If $n,k\in\mathbb N$, solve $2^8+2^{11}+2^n=k^2$.

Question

If $n,k\in\mathbb N$, solve $$2^8+2^{11}+2^n=k^2$$

It's hard for me to find an idea. Some help would be great. Thanks.

Answer 1

First assume $n\ge 8$ and write $n=m+8$, we get: $$(2^4)^2(1+2^3+2^m)=k^2$$ This is possible only if $1+2^3+2^m=9+2^m$ is a perfect square. If it is, then we can write it as $(3+p)^2$ (since it is obviously greater than $3$), and thus: $$9+6p+p^2=9+2^m$$ $$\Leftrightarrow p(6+p)=2^m$$ In particular, $p=2^r$ and $6+p=2^{m-r}$ for some $1\le r\le m$ (notice that $r=0$ is impossible). From this we get $6+p=6+2^r=2(3+2^{r-1})=2^{m-r}$, and the only solution to this equation is $r=1$, $m=4$. Thus the only solution for $n\ge8$ is $n=12$, which gives us: $$2^8+2^{11}+2^{12}=80^2$$

Now for $0\le n<8$, we have: $$2^n(2^{8-n}+2^{11-n}+1)=k^2$$ Since the number in the brackets cannot be a multiple of $2$, we observe that $n$ must be even. This leaves us with the condition that $2^{8-n}+2^{11-n}+1$ is a square. That number is exactly: $$2^{8-n}3^2+1$$ For $n=6$ we have $2^23^2+1=37$ which is not a square.

For $n=4$ we have $2^43^2+1=145$ which is not a square.

For $n=2$ we have $2^63^2+1=577$ which is again not a square.

Finally for $n=0$ we have $2^83^2+1=1153$ which is not a square either.

Thus the only solution is $n=12$.

Answer 2

HINT:

We have $\displaystyle2^8+2^{11}=2304$

Now, $2^8+2^{11}+2^0=2304+1\ne k^2$

$\displaystyle 2^8+2^{11}+2^1=2304+2\equiv2\pmod8$, but $a^2\equiv0,1,4\pmod8$

So, $n\ge2$ let $n=m+2$ where $m\ge0$

$\displaystyle 2^8+2^{11}+2^{m+2}=4(576+2^m)\implies 576+2^m$ must be perfect square

Like either method $m\ge2$ let $m=r+2$ where $r\ge0$

$\displaystyle576+2^{r+2}=4(144+2^r)$

Follow this step

One Observation : $$(2^4)^2+2\cdot2^4\cdot2^6+(2^6)^2=(2^4+2^6)^2$$

Answer 3

This is justa a partial solution.

we assume first that $n>8$ and we see $2^8+2^{11}+2^n=k^2$ as

$2^8(1+2^3+2^{n-8})=k^2$ we see $2^8$ is a square...

so the problem reduces to see when would $1+2^3+2^{n-8}$ be a square..

As we have assumed $n>8$ we would have $1+2^3+2^{n-8}$ to be an odd integer.

Now, any odd perfect square is congruent to $1$ modulo $8$

we see that $1+2^3$ is congruent to $1$ modulo $8$

so the problem is to make $2^{n-8}$ as $0$ modulo $8$

i.e., to make $n-8=3k$

least possible case would be $k=1$ i.e., $n=12$

I tried to repeat this and let $k=2$ but then It does not work.

being congruent to $1$ mod $8$ is only necessary condition but not sufficient so I can not say this works for all $k\geq 2$.

trying to solve this completely for a long time with no success. so i thought of posting what i have done and expecting for some suggestions.