$\forall\; n \geq 3 \;\;\exists (x,y) \in \{2m+1;m\in \mathbb{N}\}^2$ such that $2^n=7x^2 + y^2 $
I proved it by induction and I'm curious how you guys would you show it.
$\forall\; n \geq 3 \;\;\exists (x,y) \in \{2m+1;m\in \mathbb{N}\}^2$ such that $2^n=7x^2 + y^2 $
I proved it by induction and I'm curious how you guys would you show it.
On
If $a^2 + 7 b^2 = 2^n$ with odd $$ a \equiv b \pmod 4, $$ then we get $$ \left( \frac{a - 7b}{2} \right)^2 + 7 \left( \frac{a+b}{2} \right)^2 = 2^{n+1}, $$ we have preserved oddness as well as equality $\pmod 4,$ as $$ \left( \frac{a - 7b}{2} \right) \equiv \left( \frac{a+b}{2} \right) \pmod 4 $$ since $$ a - 7 b \equiv a + b \equiv \pm 2 \pmod 8 $$
jagy@phobeusjunior:~$
1 1 2 8
-3 1 6 16
-5 -1 2 32
1 -3 6 64
11 -1 2 128
9 5 6 256
-13 7 2 512
-31 -3 6 1024
-5 -17 2 2048
57 -11 6 4096
67 23 2 8192
-47 45 6 16384
-181 -1 2 32768
-87 -91 6 65536
275 -89 2 131072
449 93 6 262144
-101 271 2 524288
-999 85 6 1048576
-797 -457 2 2097152
1201 -627 6 4194304
2795 287 2 8388608
393 1541 6 16777216
-5197 967 2 33554432
-5983 -2115 6 67108864
4411 -4049 2 134217728
jagy@phobeusjunior:~$
By the generalized Lagrange's identity $$(a^2+7b^2)(c^2+7d^2)=(ac-7bd)^2+7(ad+bc)^2 \tag{1}$$ That can be seen as a consequence of the multiplicativity of the norm over $\mathbb{Z}[\sqrt{-7}]$.
It follows that the set $E_7$ of integers that can be represented as $n^2+7m^2$ is a semigroup.
Obviously $8=2^3=1^2+7\cdot 1^2\in E_7$, $16=3^2+7\cdot 1^2\in E_7$ and $32=5^2+7\cdot 1^2\in E_7$.
Then you may use $(1)$ and induction to prove that every integer of the form $2^n$ with $n\geq 3$ belongs to $E_7$, and it has a representation of the form $x^2+7y^2$ with both $x$ and $y$ being odd numbers.