Proving something cant be written as a square

[S.Iyengar]

p²ⁿ <= 2p^a+n- 2pⁿ - p^a + 5
Since p^a > 5:
p²ⁿ <= 2p^a+n
pⁿ <= 2p^a
Since p > 2, n <= a.

k(kp^a+2) = 2kp^a+n+2pⁿ-p²ⁿ+3
k²p^a+2k = 2kp^a+n+2pⁿ-p²ⁿ+3
So 2k congruent to 3 modulo pⁿ (all other terms are divisible by pⁿ).
2k cannot be negative, so 2k >= pⁿ+3.

Another small misunderstanding sir. You have written that , $k^2p^a+2k=2kp^{a+n}+2p^n-p^{2n}+3$ and have said that, all the other terms are divisible by $p^n$. But we can't write that, given there is a term $k^2p^a$ on the L.H.S

Thanks a lot again sir.

 

[haruspex]

Ah yes - I overlooked another place I had used the (wrong) "n <= a" result.
This looks more serious...

 

[S.Iyengar]

Ah yes - I overlooked another place I had used the (wrong) "n <= a" result.
This looks more serious...

No problem sir.. I am happy that you responded in a nice manner.