Proving something cant be written as a square

 Quote by haruspex Therefore pa divides (r+1)(r-1) Since p is a prime > 2, it cannot have factors in common with both r-1 and r+1. Therefore pa divides (r+1) or (r-1) So if $r > 1, r = kp^a \pm 1,$ some $k > 0$
Ok thank you for your infinite patience sir.

 Quote by haruspex Therefore pa divides (r+1)(r-1) Since p is a prime > 2, it cannot have factors in common with both r-1 and r+1. Therefore pa divides (r+1) or (r-1) So if $r > 1, r = kp^a \pm 1,$ some $k > 0$
Thank you again sir

 Quote by haruspex p2n <= 2pa+n- 2pn - pa + 5 Since pa > 5: p2n <= 2pa+n pn <= 2pa Since p > 2, n <= a. k(kpa+2) = 2kpa+n+2pn-p2n+3 k2pa+2k = 2kpa+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.
 Recognitions: Homework Help Science Advisor Ah yes - I overlooked another place I had used the (wrong) "n <= a" result. This looks more serious...

 Quote by haruspex 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.

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