Proving something cant be written as a square

[S.Iyengar]

My version of proof.. Some one please verify ..

I have tried this. Please do comment on my proof.

We have n^2=p^{m+a}-p^{m}+3p^a+1
So its clear that n^2 = p^{2n+2a}-p^{2n+a}+3p^a+1 ( \rm{Since} \ m=2n+a) \ [1]
n^2 \lt (p^{n+a}+p^n)^2
n^2 \gt (p^{n+a}-p^n)^2

Since

p^{2n+2a}-p^{2n+a}+3p^a+1 \lt p^{2n+2a}+p^{2n}+2p^{2n+a}
p^{2n+2a}-p^{2n+a}+3p^a+1 \gt p^{2n+2a}+p^{2n}-2p^{2n+a}

Hence if (x-z)^2 \lt y^2 \lt (x+z)^2 then we have y=(x-z+1) for some z\gt 0 .

Hence we have n=p^{n+a}-p^n+1 \implies n^2= (p^{n+a}-p^n+1)^2
n^2=p^{2n+2a}+p^{2n}+1-2p^{2n+a}-2p^n+2p^{n+a}

So it follows from [1] that p^{2n+2a}+p^{2n}+1-2p^{2n+a}-2p^n+2p^{n+a}=p^{2n+2a}-p^{2n+a}+3p^a+1
p^{2n+a}+2p^{n}+3p^{a} = p^{2n}+2p^{n+a}

So the above equation is never true as for \{\{n,a\} \gt 0 \}\in \mathbb{a} and p\gt3 the exponents on L.H.S sum up to 3n+2a where as on R.H.S sum up to 3n+a .

So p^{2n+a}+2p^{n}+3p^{a} \neq p^{2n}+2p^{n+a}.

Hence a contradiction is achieved.

 

[haruspex]

We have n^2=p^{m+a}-p^{m}+3p^a+1
So its clear that n^2 = p^{2n+2a}-p^{2n+a}+3p^a+1 ( \rm{Since} \ m=2n+a) \ [1]

You've used n to mean two different things in the same equation. Could lead to confusion.

n^2 \lt (p^{n+a}+p^n)^2
n^2 \gt (p^{n+a}-p^n)^2

Since

p^{2n+2a}-p^{2n+a}+3p^a+1 \lt p^{2n+2a}+p^{2n}+2p^{2n+a}
p^{2n+2a}-p^{2n+a}+3p^a+1 \gt p^{2n+2a}+p^{2n}-2p^{2n+a}

Hence if (x-z)^2 \lt y^2 \lt (x+z)^2 then we have y=(x-z+1) for some z\gt 0 .

No. We have y=(x-z+t) for some t, 2z\gt t\gt 0 .

Hence we have n=p^{n+a}-p^n+1 \implies n^2= (p^{n+a}-p^n+1)^2
n^2=p^{2n+2a}+p^{2n}+1-2p^{2n+a}-2p^n+2p^{n+a}

So it follows from [1] that p^{2n+2a}+p^{2n}+1-2p^{2n+a}-2p^n+2p^{n+a}=p^{2n+2a}-p^{2n+a}+3p^a+1
p^{2n+a}+2p^{n}+3p^{a} = p^{2n}+2p^{n+a}

So the above equation is never true as for \{\{n,a\} \gt 0 \}\in \mathbb{a} and p\gt3 the exponents on L.H.S sum up to 3n+2a where as on R.H.S sum up to 3n+a .

I cannot make any sense of that argument. What does \{\{n,a\} \gt 0 \}\in \mathbb{a} mean? What has the sum of the exponents to do with it?

 

[S.Iyengar]

You've used n to mean two different things in the same equation. Could lead to confusion.

No. We have y=(x-z+t) for some t, 2z\gt t\gt 0 .

I cannot make any sense of that argument. What does \{\{n,a\} \gt 0 \}\in \mathbb{a} mean? What has the sum of the exponents to do with it?

Sir sorry sir... Assume that my proof is wrong sir... Please explain me the last part of your proof sir.. As I asked in previous posts... But this time.. there is a flaw I can show in your proof sir..

r \ge p^a-1 \implies p^{2a+2n}-p^{a+2n}+3p^a+1 \le (p^{a+n} - p^a + 1)^2 .. But you wrote an opposite to that statement.. But that may be a typo sir.. I know you are seminal mathematician.. So these errors might be a result of typing..

 

[haruspex]

You're right! That is backwards. Well spotted.
So I should have obtained
p²ⁿ >= 2p^a+n- 2pⁿ - p^a + 5
This suggests n >= a mostly. E.g. suppose n = a:
p²ⁿ >= 2p²ⁿ- 2pⁿ - pⁿ + 5
p²ⁿ <= 3pⁿ - 5
pⁿ < 3
So, find where I used n <= a later and instead use n >= a. See if you can complete the proof from there.

 

[S.Iyengar]

You're right! That is backwards. Well spotted.
So I should have obtained
p²ⁿ >= 2p^a+n- 2pⁿ - p^a + 5
This suggests n >= a mostly. E.g. suppose n = a:
p²ⁿ >= 2p²ⁿ- 2pⁿ - pⁿ + 5
p²ⁿ <= 3pⁿ - 5
pⁿ < 3
So, find where I used n <= a later and instead use n >= a. See if you can complete the proof from there.

Sir please help me with this :

How will you get 3p^{a+2n} \le 6p^{a+n} + 9p^a + 4p^2n &lt; 19p^{a+n} \implies p^n \le 6 ?. More over in the expansion of (p^{a+n} - p^a + 1)^2 \implies \dfrac{p^{2a+2n}}{4} + \dfrac{3p^{2a+n}}{2} + \dfrac{9p^{2a}}{4} -p^{a+n} +3p^a. So what about the terms -p^{a+n} +3p^a ?.

 

[S.Iyengar]

You've used n to mean two different things in the same equation. Could lead to confusion.

No. We have y=(x-z+t) for some t, 2z\gt t\gt 0 .

I cannot make any sense of that argument. What does \{\{n,a\} \gt 0 \}\in \mathbb{a} mean? What has the sum of the exponents to do with it?

Its a typo sir.. They mean \{\{n,a\} \gt 0 \}\in \mathbb{Z} . I mean both of them are greater than zero and are integers.

 

[haruspex]

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

 

[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

Yes sir that's the best way. But there is a small consideration sir. we have n \ge 1 . But you considered it as n \ge 2 which will change the bounds. Thank you master

 

[haruspex]

Yes sir that's the best way. But there is a small consideration sir. we have n \ge 1 . But you considered it as n \ge 2 which will change the bounds. Thank you master

It's pretty easy to deal with the n = 1 case specifically:
3p¹⁺²ⁿ <= 6p¹⁺ⁿ + 9p + 4p^2n-a+1
3p³ <= 6p² + 9p + 4p^3-a
<= 6p² + 9p + 4p²
<= 10p² + 9p
3p² <= 10p + 9
p <= 4

 

[S.Iyengar]

It's pretty easy to deal with the n = 1 case specifically:
3p¹⁺²ⁿ <= 6p¹⁺ⁿ + 9p + 4p^2n-a+1
3p³ <= 6p² + 9p + 4p^3-a
<= 6p² + 9p + 4p²
<= 10p² + 9p
3p² <= 10p + 9
p <= 4

This time the proof looks rigorous sir. Thank you.

 

[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.