Proving something cant be written as a square

[S.Iyengar]

I think we can forget about a <= n or a >= n. All we need is a >= 1, p >= 3, n >= 2

3p^a+2n <= 6p^a+n + 9p^a + 4p²ⁿ
3p¹⁺²ⁿ <= 6p¹⁺ⁿ + 9p + 4p^2n-a+1
3p¹⁺²ⁿ <= 6p¹⁺ⁿ + 9p + 4p²ⁿ
(3p-4)p²ⁿ <= 6pⁿ⁺¹ + 9p
(3p-4)pⁿ <= 6p + 9p^1-n <= 6p + 3
Since pⁿ >= 9
(3p-4)9 <= 6p + 3
7p <= 13

What about the situation when a=k ?

 

[haruspex]

What about the situation when a=k ?

What's special about a = k? Can you point to a step where you think that matters?

 

[S.Iyengar]

What's special about a = k? Can you point to a step where you think that matters?

Dear sir,

Here you considered that

3p^{a+2n} \le 6p^{a+n} + 9p^a + 4p^{2n} \\<br /> 3p^{1+2n} \le 6p^{1+n} + 9p + 4p^{2n-a+1} \\<br /> 3p^{1+2n} \le 6p^{1+n} + 9p + 4p^{2n}\\<br /> (3p-4)p^{2n} \le 6p^{n+1} + 9p

There in the second step you have substituted a=1 and then proceeded further. So what about the case when a=k ?

Thank you sir.

 

[micromass]

No, he did not substitute a=1.

He multiplied both sides of the inequality by something. Can you figure it out now?

 

[S.Iyengar]

No, he did not substitute a=1.

He multiplied both sides of the inequality by something. Can you figure it out now?

Thank you sir.. I think he multiplied both sides with p^{-a+1} . Thanks a lot

 

[haruspex]

Dear sir,

Here you considered that

3p^{a+2n} \le 6p^{a+n} + 9p^a + 4p^{2n} \\<br /> 3p^{1+2n} \le 6p^{1+n} + 9p + 4p^{2n-a+1} \\<br /> 3p^{1+2n} \le 6p^{1+n} + 9p + 4p^{2n}\\<br /> (3p-4)p^{2n} \le 6p^{n+1} + 9p

There in the second step you have substituted a=1 and then proceeded further. So what about the case when a=k ?

Thank you sir.

No, it's not a question of substituting a = 1 as such. We are assuming a >= 1, so 4p^{2n-a+1} &lt;= 4p^{2n}

 

[S.Iyengar]

No, it's not a question of substituting a = 1 as such. We are assuming a >= 1, so 4p^{2n-a+1} &lt;= 4p^{2n}

But I find another flaw sir.

You wrote that

r^2-1 = p^a(2rp^n-p^{2n}+3) = (r+1)(r-1) [2]

So if r &gt; 1, r = kp^a \pm 1, some k &gt; 0 [3]

So the fact that [2] \implies [3] is not correct. Because it follows from the argument that either (r-1) \rm{or} \ (r+1) | (2rp^n-p^{2n}+3).

So the best counter example is 8*6=48 = 2^4*3. Then it doesn't mean that either 8 \ \rm{or} \ 6 | 3. So its not correct, and thereby the entire proof collapse.

I whole-heartedly apologize if my argument is wrong.

 

[haruspex]

r^2-1 = p^a(2rp^n-p^{2n}+3) = (r+1)(r-1) [2]

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 r &gt; 1, r = kp^a \pm 1, some k &gt; 0

 

[S.Iyengar]

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 r &gt; 1, r = kp^a \pm 1, some k &gt; 0

Ok thank you for your infinite patience sir.

 

[S.Iyengar]

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 r &gt; 1, r = kp^a \pm 1, some k &gt; 0

Thank you again sir

 

[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.