Proving something cant be written as a square

[haruspex]

$d.(n+1)-3= p^t(1-p^{-\phi})$---(3)

So here it's evident that $p^t | d.(n+1)-3$ .

No, you can't do that. $1-p^{-\phi}$ is not an integer.
E.g. 18 = 3³-3² = 3³(1-3^-1), but 3³ does not divide 18.

 

[Norwegian]

Strange thread indeed. Have you agreed on what statement you are looking to prove? For example, when p=3, m=a+2 then (p^m+3)(p^a-1)+4 is a square, so you may want p>3 in your original statement.

 

[S.Iyengar]

Strange thread indeed. Have you agreed on what statement you are looking to prove? For example, when p=3, m=a+2 then (p^m+3)(p^a-1)+4 is a square, so you may want p>3 in your original statement.

Yes sir, I am sorry. There p>3 actually. But later I came to know that exception

 

[S.Iyengar]

No, you can't do that. $1-p^{-\phi}$ is not an integer.
E.g. 18 = 3³-3² = 3³(1-3^-1), but 3³ does not divide 18.

So is there any alternate way of fixing it sir ?. I do know , but I thought that the equation would imply that.

 

[S.Iyengar]

Strange thread indeed. Have you agreed on what statement you are looking to prove? For example, when p=3, m=a+2 then (p^m+3)(p^a-1)+4 is a square, so you may want p>3 in your original statement.

Sir, I am sorry, can you suggest me some hints to prove that statement sir. I do apologize for not mentioning that special exception which you pointed out cleverly.

 

[S.Iyengar]

I did another attempt of solving it. Let me post it here and wait for comments. I think this time I didnt make any mistake.

 

[haruspex]

Put (p^a+n - r)² = (p^a+2n+3)(p^a-1)+4 = p^2a+2n-p^a+2n+3p^a+1 [1]
r²-1 = p^a(2rpⁿ-p²ⁿ+3) = (r+1)(r-1) [2]
So if r > 1, r = kp^a+/-1, some k > 0 [3]
In particular, r >= p^a-1
From [1]
p^2a+2n-p^a+2n+3p^a+1 >= (p^a+n - p^a + 1)²
p^2a+2n-p^a+2n+3p^a+1 >= p^2a+2n - 2p^2a+n + 2p^a+n + p^2a - 2p^a + 1
-p²ⁿ+3 >= - 2p^a+n + 2pⁿ + p^a - 2
p²ⁿ <= 2p^a+n- 2pⁿ - p^a + 5
Hence n <= a
Returning to [3], consider r = kp^a+1. From [2]:
k(kp^a+2) = 2kp^a+n+2pⁿ-p²ⁿ+3
2k >= pⁿ+3
Similarly if r = kp^a-1 then 2k >= pⁿ-3
So r >= p^a+n/2 - 3p^a/2 - 1
From [1]
p^2a+2n-p^a+2n+3p^a+1 <= (p^a+n - p^a+n/2 + 3p^a/2 - 1)²
p^2a+2n-p^a+2n+3p^a+1 <= (p^a+n/2 + 3p^a/2 - 1)²
p^2a+2n-p^a+2n+3p^a <= p^2a+2n/4 + 3p^2a+n/2 + 9p^2a/4
3p^2a+2n/4 <= 3p^2a+n/2 + 9p^2a/4 + p^a+2n
3p^a+2n <= 6p^a+n + 9p^a + 4p²ⁿ < 19p^a+n
pⁿ <= 6
It shouldn't be hard from there.

 

[S.Iyengar]

Put (p^a+n - r)² = (p^a+2n+3)(p^a-1)+4 = p^2a+2n-p^a+2n+3p^a+1 [1]
r²-1 = p^a(2rpⁿ-p²ⁿ+3) = (r+1)(r-1) [2]
So if r > 1, r = kp^a+/-1, some k > 0 [3]
In particular, r >= p^a-1
From [1]
p^2a+2n-p^a+2n+3p^a+1 >= (p^a+n - p^a + 1)²
p^2a+2n-p^a+2n+3p^a+1 >= p^2a+2n - 2p^2a+n + 2p^a+n + p^2a - 2p^a + 1
-p²ⁿ+3 >= - 2p^a+n + 2pⁿ + p^a - 2
p²ⁿ <= 2p^a+n- 2pⁿ - p^a + 5
Hence n <= a
Returning to [3], consider r = kp^a+1. From [2]:
k(kp^a+2) = 2kp^a+n+2pⁿ-p²ⁿ+3
2k >= pⁿ+3
Similarly if r = kp^a-1 then 2k >= pⁿ-3
So r >= p^a+n/2 - 3p^a/2 - 1
From [1]
p^2a+2n-p^a+2n+3p^a+1 <= (p^a+n - p^a+n/2 + 3p^a/2 - 1)²
p^2a+2n-p^a+2n+3p^a+1 <= (p^a+n/2 + 3p^a/2 - 1)²
p^2a+2n-p^a+2n+3p^a <= p^2a+2n/4 + 3p^2a+n/2 + 9p^2a/4
3p^2a+2n/4 <= 3p^2a+n/2 + 9p^2a/4 + p^a+2n
3p^a+2n <= 6p^a+n + 9p^a + 4p²ⁿ < 19p^a+n
pⁿ <= 6
It shouldn't be hard from there.

Sir, thanks a lot for your answer. But the answer was so short and I can't understand the following points sir.

* How can we know that n<=a from p^2n <= 2p^{a+n}- 2p^n - p^a + 5 ? .

* Similarly How do we get 2k>=p^n+3 from this one sir k(kp^a+2) = 2kp^{a+n}+2p^n-p^{2n}+3?

I will be thankful if you can answer this

 

[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.

 

[S.Iyengar]

Put (p^a+n - r)² = (p^a+2n+3)(p^a-1)+4 = p^2a+2n-p^a+2n+3p^a+1 [1]
r²-1 = p^a(2rpⁿ-p²ⁿ+3) = (r+1)(r-1) [2]
So if r > 1, r = kp^a+/-1, some k > 0 [3]
In particular, r >= p^a-1
From [1]
p^2a+2n-p^a+2n+3p^a+1 >= (p^a+n - p^a + 1)²
p^2a+2n-p^a+2n+3p^a+1 >= p^2a+2n - 2p^2a+n + 2p^a+n + p^2a - 2p^a + 1
-p²ⁿ+3 >= - 2p^a+n + 2pⁿ + p^a - 2
p²ⁿ <= 2p^a+n- 2pⁿ - p^a + 5
Hence n <= a
Returning to [3], consider r = kp^a+1. From [2]:
k(kp^a+2) = 2kp^a+n+2pⁿ-p²ⁿ+3
2k >= pⁿ+3
Similarly if r = kp^a-1 then 2k >= pⁿ-3
So r >= p^a+n/2 - 3p^a/2 - 1
From [1]
p^2a+2n-p^a+2n+3p^a+1 <= (p^a+n - p^a+n/2 + 3p^a/2 - 1)²
p^2a+2n-p^a+2n+3p^a+1 <= (p^a+n/2 + 3p^a/2 - 1)²
p^2a+2n-p^a+2n+3p^a <= p^2a+2n/4 + 3p^2a+n/2 + 9p^2a/4
3p^2a+2n/4 <= 3p^2a+n/2 + 9p^2a/4 + p^a+2n
3p^a+2n <= 6p^a+n + 9p^a + 4p²ⁿ < 19p^a+n
pⁿ <= 6
It shouldn't be hard from there.

Sir, thanks a lot for your patient replies. I have one final doubt sir, Why did you take $(p^{a+n}-r)^2=(p^{a+2n}+3)(p^a-1)+4 = p^{2a+2n}-p{a+2n}+3p^a+1$ ?. I know we need to equate the term to some perfect square. But the perfect square would be a square of even number always as we have $+1$ on the extreme end. So do you know that $p^{a+n}-r$ is an even number ?. If so $r$ should be odd to force the difference to be even. But how do you know that square on the right is of the form $(p^{a+n}-r)^2$.

Thanks a lot for your patience sir. Please do help me with this last one sir.

 

[S.Iyengar]

Put (p^a+n - r)² = (p^a+2n+3)(p^a-1)+4 = p^2a+2n-p^a+2n+3p^a+1 [1]
r²-1 = p^a(2rpⁿ-p²ⁿ+3) = (r+1)(r-1) [2]
So if r > 1, r = kp^a+/-1, some k > 0 [3]
In particular, r >= p^a-1
From [1]
p^2a+2n-p^a+2n+3p^a+1 >= (p^a+n - p^a + 1)²
p^2a+2n-p^a+2n+3p^a+1 >= p^2a+2n - 2p^2a+n + 2p^a+n + p^2a - 2p^a + 1
-p²ⁿ+3 >= - 2p^a+n + 2pⁿ + p^a - 2
p²ⁿ <= 2p^a+n- 2pⁿ - p^a + 5
Hence n <= a
Returning to [3], consider r = kp^a+1. From [2]:
k(kp^a+2) = 2kp^a+n+2pⁿ-p²ⁿ+3
2k >= pⁿ+3
Similarly if r = kp^a-1 then 2k >= pⁿ-3
So r >= p^a+n/2 - 3p^a/2 - 1
From [1]
p^2a+2n-p^a+2n+3p^a+1 <= (p^a+n - p^a+n/2 + 3p^a/2 - 1)²
p^2a+2n-p^a+2n+3p^a+1 <= (p^a+n/2 + 3p^a/2 - 1)²
p^2a+2n-p^a+2n+3p^a <= p^2a+2n/4 + 3p^2a+n/2 + 9p^2a/4
3p^2a+2n/4 <= 3p^2a+n/2 + 9p^2a/4 + p^a+2n
3p^a+2n <= 6p^a+n + 9p^a + 4p²ⁿ < 19p^a+n
pⁿ <= 6
It shouldn't be hard from there.

And we get another contradiction sir. That is $(p^{a+n}-r)$ should be an even number because the term on the R.H.S is even always. So its clear that $p^{a+n}$ is odd so $r$ should be odd so as to push the difference to be even. If that is the case according to [2] $r^2-1= p^a(2rp^n-p^{2n}+3)$ the L.H.S would be even ( since $r^2-1$ as $r$ is odd. But the R.H.S is $p^a(2rp^n-p^{2n}+3)$ . $2rp^n-p^{2n}$ would be even. Then $2rp^n-p^{2n}+3$ would be odd. $p^a(2rp^n-p^{2n}+3)$ is again odd . So its a clear contradiction if we assume that $r$ is odd, then L.H.S $\neq$ R.H.S . In case if don't assume that $r$ is odd , the [1] don't make sense as the R.H.S should be even.

Please do clarify this sir.

 

[haruspex]

Sir, thanks a lot for your patient replies. I have one final doubt sir, Why did you take $(p^{a+n}-r)^2=(p^{a+2n}+3)(p^a-1)+4 = p^{2a+2n}-p{a+2n}+3p^a+1$ ?. I know we need to equate the term to some perfect square. But the perfect square would be a square of even number always as we have $+1$ on the extreme end. So do you know that $p^{a+n}-r$ is an even number ?. If so $r$ should be odd to force the difference to be even. But how do you know that square on the right is of the form $(p^{a+n}-r)^2$.

Thanks a lot for your patience sir. Please do help me with this last one sir.

Yes r, if it exists at all, will necessarily be odd. But I don't see why you think that's a problem. At no point did I assume it to be even.

 

[S.Iyengar]

Yes r, if it exists at all, will necessarily be odd. But I don't see why you think that's a problem. At no point did I assume it to be even.

Yes sir, I didn't told that you have assumed r to be even. I just said that there would be a contradiction even if we assume r to be odd.

But why is the term $(p^{a+n}-r)^2$ taken there sir ?.

 

[haruspex]

But why is the term $(p^{a+n}-r)^2$ taken there sir ?.

Because the leading term on the RHS is the square of p^a+n. So it seemed likely that writing it as the square of some offset from there would allow a lot of cancellation. That makes it easier to see where to go next.

 

[S.Iyengar]

Because the leading term on the RHS is the square of p^a+n. So it seemed likely that writing it as the square of some offset from there would allow a lot of cancellation. That makes it easier to see where to go next.

Sir a nice answer sir. But your answer serves only in showing that $(p^m+3)(p^a-1)+4$ is not a square of $(p^{a+n}-r)$. But in general it doesn't hold for some other $n \neq (p^{a+n}-r)$.

I mean proving that there is a contradiction for some $(p^{a+n}-r)$ will not hold good for other numbers. Your proof means that $(p^m+3)(p^a-1)+4$ will never be a square of the form $(p^{a+n}-r)$. It doesn't mean that $(p^m+3)(p^a-1)+4$ is not a square. There is a chance that it can be a square of other numbers.

Please do clarify this doubt sir.

 

[S.Iyengar]

Put (p^a+n - r)² = (p^a+2n+3)(p^a-1)+4 = p^2a+2n-p^a+2n+3p^a+1 [1]
r²-1 = p^a(2rpⁿ-p²ⁿ+3) = (r+1)(r-1) [2]
So if r > 1, r = kp^a+/-1, some k > 0 [3]
In particular, r >= p^a-1
From [1]
p^2a+2n-p^a+2n+3p^a+1 >= (p^a+n - p^a + 1)²
p^2a+2n-p^a+2n+3p^a+1 >= p^2a+2n - 2p^2a+n + 2p^a+n + p^2a - 2p^a + 1
-p²ⁿ+3 >= - 2p^a+n + 2pⁿ + p^a - 2
p²ⁿ <= 2p^a+n- 2pⁿ - p^a + 5
Hence n <= a
Returning to [3], consider r = kp^a+1. From [2]:
k(kp^a+2) = 2kp^a+n+2pⁿ-p²ⁿ+3
2k >= pⁿ+3
Similarly if r = kp^a-1 then 2k >= pⁿ-3
So r >= p^a+n/2 - 3p^a/2 - 1
From [1]
p^2a+2n-p^a+2n+3p^a+1 <= (p^a+n - p^a+n/2 + 3p^a/2 - 1)²
p^2a+2n-p^a+2n+3p^a+1 <= (p^a+n/2 + 3p^a/2 - 1)²
p^2a+2n-p^a+2n+3p^a <= p^2a+2n/4 + 3p^2a+n/2 + 9p^2a/4
3p^2a+2n/4 <= 3p^2a+n/2 + 9p^2a/4 + p^a+2n
3p^a+2n <= 6p^a+n + 9p^a + 4p²ⁿ < 19p^a+n
pⁿ <= 6
It shouldn't be hard from there.

Very sorry to say this sir, another irreparable mistake if $r \ge \dfrac{p^{a+n}}{2} - \dfrac{3p^a}{2} - 1 \implies p^{2a+2n}-p^{a+2n}+3p^a+1 \ge (p^{a+n} - \dfrac{p^{a+n}}{2} - \dfrac{3p^a}{2} - 1 )^2$. You wrote complete inverse of it..

 

[haruspex]

Sir a nice answer sir. But your answer serves only in showing that $(p^m+3)(p^a-1)+4$ is not a square of $(p^{a+n}-r)$. But in general it doesn't hold for some other $n \neq (p^{a+n}-r)$.

I mean proving that there is a contradiction for some $(p^{a+n}-r)$ will not hold good for other numbers. Your proof means that $(p^m+3)(p^a-1)+4$ will never be a square of the form $(p^{a+n}-r)$. It doesn't mean that $(p^m+3)(p^a-1)+4$ is not a square. There is a chance that it can be a square of other numbers.

Please do clarify this doubt sir.

If it is the square of k then define r = p^a+n-k.

 

[haruspex]

Very sorry to say this sir, another irreparable mistake if $r \ge \dfrac{p^{a+n}}{2} - \dfrac{3p^a}{2} - 1 \implies p^{2a+2n}-p^{a+2n}+3p^a+1 \ge (p^{a+n} - \dfrac{p^{a+n}}{2} - \dfrac{3p^a}{2} - 1 )^2$. You wrote complete inverse of it..

No, it's right. We have (X-r)2 = RHS, X >=r >= 0. We've shown r >= Y. When r = Y we'll get the max possible value of the RHS. When r is larger it will make the RHS smaller.
So RHS <= (X-Y)².

 

[S.Iyengar]

No, it's right. We have (X-r)2 = RHS, X >=r >= 0. We've shown r >= Y. When r = Y we'll get the max possible value of the RHS. When r is larger it will make the RHS smaller.
So RHS <= (X-Y)².

I am extremely sorry for being stupid. I now take back my words literally. I learned that we must have a strong proof to say such sentences like "irreparable mistake" etc.. Or else they would be very much foolish.

 

[S.Iyengar]

Put (p^a+n - r)² = (p^a+2n+3)(p^a-1)+4 = p^2a+2n-p^a+2n+3p^a+1 [1]
r²-1 = p^a(2rpⁿ-p²ⁿ+3) = (r+1)(r-1) [2]
So if r > 1, r = kp^a+/-1, some k > 0 [3]
In particular, r >= p^a-1
From [1]
p^2a+2n-p^a+2n+3p^a+1 >= (p^a+n - p^a + 1)²
p^2a+2n-p^a+2n+3p^a+1 >= p^2a+2n - 2p^2a+n + 2p^a+n + p^2a - 2p^a + 1
-p²ⁿ+3 >= - 2p^a+n + 2pⁿ + p^a - 2
p²ⁿ <= 2p^a+n- 2pⁿ - p^a + 5
Hence n <= a
Returning to [3], consider r = kp^a+1. From [2]:
k(kp^a+2) = 2kp^a+n+2pⁿ-p²ⁿ+3
2k >= pⁿ+3
Similarly if r = kp^a-1 then 2k >= pⁿ-3
So r >= p^a+n/2 - 3p^a/2 - 1
From [1]
p^2a+2n-p^a+2n+3p^a+1 <= (p^a+n - p^a+n/2 + 3p^a/2 - 1)²
p^2a+2n-p^a+2n+3p^a+1 <= (p^a+n/2 + 3p^a/2 - 1)²
p^2a+2n-p^a+2n+3p^a <= p^2a+2n/4 + 3p^2a+n/2 + 9p^2a/4
3p^2a+2n/4 <= 3p^2a+n/2 + 9p^2a/4 + p^a+2n
3p^a+2n <= 6p^a+n + 9p^a + 4p²ⁿ < 19p^a+n
pⁿ <= 6
It shouldn't be hard from there.

Sir one final doubt, after this I won't again raise any doubts. My brain is too small to grasp the direct simplifications. I have tried a lot to obtain the same result but can't do it. How will you get $3p^{a+2n} \le 6p^{a+n} + 9p^a + 4p^2n < 19p^{a+n} \implies p^n \le 6$ ?. 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 $-p^{a+n} +3p^a$ ?.

The expression becomes

$p^{2a+2n}-p^{a+2n}+3p^a \le \dfrac{p^{2a+2n}}{4} + \dfrac{3p^{2a+n}}{2} + \dfrac{9p^{2a}}{4} -p^{a+n} +3p^a$. So Please do explain that sir.

Sometimes I can understand that you loose your patience. But I tried a lot myself in understanding that, but couldn't do . Thanks a lot sir.

 

[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 $3p^{a+2n} \le 6p^{a+n} + 9p^a + 4p^2n < 19p^{a+n} \implies p^n \le 6$ ?. 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.