So, the problem written in the title is what I have in front of me. On seeing the problem, my first shot was to write $31$ as $2^5 - 1$. On expanding further, this is what the expression finally reduced to -

$$ 2^{100}(1 + 2^{92} - 2^{97} + 2^{n-100}) $$

Since $2^{100}$ is a perfect square , we may leave it aside and focus only on the part inside the parenthesis. From here on, I can't see any way out.
Taking a cue from If $2^{2017} + 2^{2014} + 2^n$ is a perfect square, find $n$., I decided to write the expression as $$ (2^{50} - x)^2$$

which opens as $$2^{100} - 2^{51}x + x^2$$

After this I am looking to compare the respective terms but I am not really getting how to go about it. Any help or any new approach would be appreciated.



There is no solution.

First, $\color{red}{n\gt196}$ since $2^n+2^{100}\gt 31\cdot2^{192}\Rightarrow 2^{n-192}+\dfrac{1}{2^{92}}\gt31$.

It follows that $1-31\cdot2^{92}+2^{n-100}$ is a natural integer.

If the claim of the post is true, then $1-31\cdot2^{92}+2^{n-100}=(2^ay+1)^2$ with $y$ odd implies $$2^{92}(2^{n-192}-31)=2^{a+1}y(2^{a-1}y+1)\Rightarrow a=91$$ so $$2^{n-192}=2^{90}y^2+y+31$$ Put $y=2^by_1+1$ with $y_1$ odd so $$2^{n-192}=2^{90}(2^by_1+1)^2+2^by_1+2^5$$

$b\gt5$ and $b\lt5$ are impossible because both give absurdes (even equal odd and integer equal non-integer, respectively). Let $b=5$ so one has $$2^{n-197}=2^{85}(2^5y_1+1)^2+y_1+1$$ Put $y_1=2^cy_2+1$ where $y_2$ is odd. Then $c=1$ if we want to avoid absurdes and the iteration of the procedure needs to do equal to $1$ all the successive exponents $c_i$ in $y_i=2^{c_i}y_{i-1}+1$ until the factor $2^{85}$ disappears.

Meanwhile if $n\lt 197+85=282$ then we could have "fraction = integer", absurde so we need $\color{red}{n\ge 282}$.

At the end we have the expression $$2^N=(2M+1)^2+2=4M^2+4M+3$$ in which if $N\gt0$ then "even =odd" and if $N=0$ then the equation $2M^2+2M+1=0$ without real solution.


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