2024/005) For m > 1 show that if $2^{2m+1} > n^2$ then $2^{2m+1} \ge n^2+7$

Because $m > 1$ $2m + 1\ge 4$  So    $2^{2m+1}$ is divisible by 16.

  $2^{2m+1}$ is double of $(2^m)^2$ and is not a square.

Now we consider 2 cases

Let us consider when n is odd 

Then $n= (2k+1)$

So $n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 4k(k+1)+1$

Ss k or k+1 is even we $n^2 \equiv 1 \pmod 8$ 

So $2^{2m+1}$ is divisible by 8 and minimum number above $n^2$ which is divisible by 8 is $n^2 + 7$

So we have proved for the case n is odd.

Let us consider when n is even

If  $n^2$ is divisible by 8 then $n^2$ is divisible by 16 and as $2^{2m+1}$ is divisible by 16 and n^2 is divisible by 16 

So $2^{2m+1}-n^2 = 16k$ for some positive k

So $2^{2m+1}-n^2 \ge16$

Hence  $2^{2m+1}-n^2 \gt 7$ or $2^{2m+1} \gt n^2+ 7$

If $n^2$ is even and is not divisible by 8 then n is of the form 4k+2

$n^2= 16k^2 + 16k + 4$

Or $n^2 \equiv 4 \pmod {16}$ 

 As $2^{2m+1} = 0 \pmod {16}$

So $2^{2m+1} \ge n^2 + 12$

Hence  $2^{2m+1} \gt n^2 + 7$

 we have proved for all 3 cases hence done