Given two non-negative integer numbers $m$ and $n$ such that $m>1$ and $2^{2m+1}\ge n^2$

Question

Given two non-negative integer numbers $m$ and $n$ such that $m>1$ and $2^{2m+1}\ge n^2$ Show that $2^{2m+1}$ $\ge n^2+7$

My idea is to proof with upper bound with assumtion n is greatest possibility

Answer 1

$n^2\equiv0,1 \pmod 4$ so there are only two ways to produce a counter example. $2^{2m+1}=n^2+3$ and $2^{2m+1}=n^2+4$

In the first case $n$ must be odd so $\exists k\in \Bbb{N} \mid 2k+1=n$ $$2^{2m+1}=(2k+1)^2+3$$ $$2^{2m+1}=4k^2+4k+4$$ $$2^{2m-1}=k^2+k+1$$ $k^2+k+1\equiv1,3 \pmod 4$, but $2^{2m-1}\equiv0 \pmod 4$ this is impossible.

In the second case $n$ must be even so $\exists k\in \Bbb{N} \mid 2k=n$ $$2^{2m+1}=(2k)^2+4$$ $$2^{2m+1}=4k^2+4$$ $$2^{2m-1}=k^2+1$$ $k^2+1\equiv1,2 \pmod 4$, but $2^{2m-1}\equiv0 \pmod 4$ this is impossible.

Answer 2

Idea: Show with induction over $m\ge 2$ that $$ 2^{2m+1}-\left( \biggl\lfloor \sqrt{2^{2m+1}\biggr\rfloor}\right)^2\ge 7. $$

Answer 3

Make use of reductio ad absurdum. Suppose that $m$ and $n$ be positive integers such that $m \ge 2$, $2^{2m+1} \ge n^2$, but $2^{2m+1} < n^2+7$. Then $2^{2m+1}-6 \le n^2 \le 2^{2m+1}$. Therefore, the following Diophantine equation has a solution. $$ 2^{2m+1}-n^2 = r , \qquad m \ge 2 , \qquad r=0,1,2,3,4,5,6 $$ Since $2m+1$ is odd, we can exclude $r=0$. Viewing in modulo $4$, we can exclude $r=1,2,5,6$. Thus we have $r=3,4$. Suppose that $r=3$, then we have $-1 \equiv n^2 \pmod{ 3}$ which is an absurdity. Thus we have the only possibility; $2^{2m+1} -n^2 = 4$. We should have $n = 2k$. Substituting yields $2^{2m-1} -k^2 =1$. Since $m \ge 2$, taking residue class modulo $4$ again, we have $k^2 \equiv -1 \pmod{4}$ which leads to a contradiction and hence, completes the proof.