Proving something cant be written as a square

S.Iyengar

Quote by haruspex

Therefore p^a 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 p^a divides (r+1) or (r-1)
So if [itex]r > 1, r = kp^a \pm 1,[/itex] some [itex] k > 0[/itex]

Ok thank you for your infinite patience sir.

S.Iyengar

Quote by haruspex

Therefore p^a 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 p^a divides (r+1) or (r-1)
So if [itex]r > 1, r = kp^a \pm 1,[/itex] some [itex] k > 0[/itex]

Thank you again sir

S.Iyengar

Quote by haruspex

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 , [itex]k^2p^a+2k=2kp^{a+n}+2p^n-p^{2n}+3[/itex] and have said that, all the other terms are divisible by [itex] p^n[/itex]. But we can't write that, given there is a term [itex] k^2p^a[/itex] 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

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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haruspex Recognitions: Homework Help Science Advisor	Ah yes - I overlooked another place I had used the (wrong) "n <= a" result. This looks more serious...