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\title{Sophie Germain Diophantine Equation}
\author{Amit Yaron}
\date{September 23, 2025}
\maketitle
When one talks about Sophie Germain, they usually take about the factorization
of $a^4+4b^4$. Let us factorize it:
\begin{align*}
  a^4+4b^4&={(a^2)}^2+{(2b^2)}^2\\
  &={(a^2)}^2+2a^2\cdot2b^2+{(2b^2)}^2-2a^2\cdot2b^2\\
  &={(a^2+2b^2)}^2-2^2a^2b^2\\
  &={(a^2+2b^2)}^2-(2ab)^2\\
  &=(a^2+2b^2-2ab)(a^2+2b^2+2ab)\\
  &=(a^2-2ab+b^2+b^2)(a^2+2ab+b^2+b^2)\\
  &=[(a-b)^2+b^2][(a+b)^2+b^2]
\end{align*}
You will probably want to use that factorization, when you have to find
values for which an expression yields a prime number, but modular arithmetic
will not help you.
For example, one I've found on Michael Penn's YouTube channel:\\
For what values of $x$ 
\[
  x^8+2^{2^x+2}
\]
is a prime number?.
Let us try some small numbers and prove that other solutions do not 
exist.
\[
  x=0\Rightarrow x^8+2^{2^x+2}=2^{2^0+2}=2^3=8
\]
Not a prime.
\[
  x=1\Rightarrow x^8+2^{2^x+2}=1^8+2^{2^1+2}=1+2^4=1+16=17
\]
17 is a prime number, so $x=1$ is a solution.\\ \\
And if $x<0$, then $2^x+2$ is a rational fraction and $2$ is a prime number,
thus:
\[
  2^{2^x+2}\notin \mathbb{Z}
\]

For values $x$ larger than or equal to 2, $2^x$ is a multiple of 4. 
Thus, we can use:
\[
  x^8+2^{2^x+2}={(x^2)}^4+2^{2^x}\cdot2^2={(x^2)}^4+4\cdot2^{2^x}
\]
Let us substitute:
\[
  a=x^2,\ b=2^{2^{\frac{x}4}}=2^{2^{x-2}}
\]
Then, 
\[
  a^4+4b^4=[(a-b)^2+b^2][(a+b)^2-b^2]
\]
is a product of 2 factors, and for that product to be prime one of the
factors has to be 1. Each factors is a sum of two squares, and squares
are non-negative. Let us check the value of $b$ when $x\geq2$"
\[
  b=2^{2^{x-2}}\geq2^{2^0}=2^1=2>1
\]
which means that 
\[
  (a\pm b)^2\geq0\land b>1\Rightarrow (a\pm b)^2+b^2\neq 1
\]
None of the factors is 1, thus for $x>2$ there are no solution, and thus:
\[
  x=1
\]
is the only solution.
\end{document}