Find all $x$ such that $4^{27}+4^{1000}+4^{x}$ is a perfect square.

Question

Let,$4^{27}+4^{1000}+4^{x}=n^2$

$4^{2}(4^{25}+4^{998}+4^{x-2})=n^2$

LHS is multiple of 4.$n$ is also multiple of 4

Let $n=4a _1$.

$4^{2}(4^{25}+4^{998}+4^{x-2})=16.a_1^2$

$4^{25}+4^{998}+4^{x-2}=4.a_1^2$

It implies $4^{x-2}$ is also multiple of 4 . Therefore $x-4$ Is natural number.

$4^2(4^{23}+4^{994}+4^{x-4})=4.a_1^2$

$4^{23}+4^{996}+4^{x-4}=a_1^2$

LHS is multiple of 4.$a_1$ is also multiple of 4

Let $a_1=4a _2$.

Similarly , we can find infinitely many integers $a_1\gt a_2\gt a_3\gt ...\gt a_n$ that satisfies this equation. By using infinite decent we prove it has no solution.

It is fake proof. We can find natural number $n=1972$ such that $4^{27}+4^{1000}+4^{1972}$ is a perfect square.

Did you find which step is wrong?

Answer 1

We can write the expression as $4^{27}(1+4^{973}+4^{x-27})$, assuming $x>27$. Then $4^{x-27} = (2^{x-27})^2$ and the next biggest square is $(2^{x-27}+1)^2 = 4^{x-27}+2^{x-26}+1.$

Certainly $4^{27}$ is a square. In order for $1+4^{973}+4^{x-27}$ to be a square, it must be larger than or equal to $4^{x-27}+2^{x-26}+1.$ So we have

$$2^{x-27} +1 \leq 4^{973} + 1$$

which gives us $x\leq 1972$.

So we know there are only finitely many solutions.

Similarly, $4^{973}= (2^{973})^2$ and the next biggest square is $(2^{973}+1)^2 = 4{973}+2^{974}+1$, so we must have

$$4{973}+2^{974}+1 \leq 4^{973}+4^{x-27}+1$$

if the latter is to be a square. This inequality gives $x\geq 469.$

So now we "just" have to check every value from $x=496$ to $1972.$

We'll leave that for interested scholar...

Edit: I ran a quit Maple loop and found $x=514$ and $x=1972$ as the only solutions.

Answer 2

There are quite a lot of mistakes in this argument, but most of them are relatively minor and can be fixed. The one mistake that can't be fixed and makes the argument completely doomed is that you assume the exponents on the left side of the equation are always positive, so you can keep factoring out $4$s. This is not true, since every time you divide by $4$ the exponents decrease, and one of them will eventually hit $0$. So you can only repeat your descent a finite number of times.

(In fact, you might run into trouble at the very first step: if $x=0$, you can't even say the initial $n$ is divisible by $4$. But even if you assume $x$ is large, the argument will collapse when the $4^{27}$ term has been reduced to $4^0$.)

Answer 3

For each iterations the powers decrease. So the claim that we can reiterate infinitely obviously false and the proof fails.

Now if $4^{1000} + 4^{27} + 4^x = n^2$ then $n^2 > 4^{1000}$ and $n > 2^{1000}$.

Let $n = 2^{1000}+k$ ($k > 0$) then $n^2 = (k + 2^{1000})^2 = 4^{1000} + 2^{1001}k + k^2 = 4^{1000} + 4^{27} + 4^x$ and

$4^{500}*2k + k^2 = 4^{27} + 4^x$

$4^x + 4^{27} = 4^{500}*2k + k^2 > 4^{500}$ so $x \ge 500$.

Which mean $4^{27}|k^2$ or $2^{27}|k$. Let $k = 2^{27}j$ then

$4^{x - 27} + 1 = 4^{473}*2^{28}j + j^2=4^{487}*j + j^2$

If $j = 1$ we have our first solution. $x = 487 + 27 =514$ and $n = 2^{1000} + k = 2^{1000} +2^{27}$

Then $(2^{1000} + 2^{27})^2 = 4^{1000} + 2*2^{1000 + 27} + 4^{27} = 4^{1000} + 4^{514} + 4^{27}$.

What about other solutions? Clearly $j$ is odd.

I'm going to quit while I'm ahead. I continued in this manner and got that there where no more solutions but clearly I made an arithmetic mistake.

What I want to od is set $j = 2m + 1$ so

$4^{x - 27} + 1 =4^{487}*j + j^2= 4^{487}2m + 4^{487} + 4m^2 + 4m + 1$

$4^{x-28} =4^{486}2m + 4^{486} + m^2 + m$ so $m$ is a mmultiple of $4^{486}$ and divide but ... well, I make some error when I try that.