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FINAL EDIT : Prove that if $p^z|n^2-1$

$p^{x-z}(p^{z}-1)=\dfrac{ n^2-1}{p^z}-3$ doesn't hold for any chosen values of $p,x,n$ and $z$.

Here $p>3$ is an odd prime , $x=2y+z, \ \{\{x,y,z\}>0\} \in \mathbb{Z}$ . There $n$ is an even number.

If the above statement is prove it will lead to a contradiction$^*$


$^*$: to understand the contradiction you need to read this :

EDIT : [History] : If anybody remembered the previous question of mine, I asked to prove $(p^x+3)(p^z-1)+4$ is not a perfect square. So I tried this $p^{x+z}-p^x+3p^z+1=l^2$ $p^{x+z}-p^x+3p^{z}=l^2-1$ $p^z(p^{x}-p^{x-z}+3)=(l+1)(l-1)$

Here its evident that $p^z$ divides either of $(l-1)$ or $(l+1)$. So let us assume as case (i) the $p^z|(l-1)$.

So let $k=\large \frac{l-1}{p^z}$, so when $k$ is decimal ( clearly $l-1$ is odd and $p^z$ is odd, so all the time the $k$ is not a integer ) ignore the case as it leads to a contradiction and proves it. Now look when $k$ is integer so the equation $p^z(p^{x}-p^{x-z}+3)=(l+1)(l-1)$ can be written as $(p^{x}-p^{x-z}+3)=k(l+1)$ $p^{x}-p^{x-z}=k(l+1)-3.$ After working on many examples, I have found an interesting pattern between the differences between the same odd number raised to different powers. For suppose we take an odd number $5$ and then work on the difference of the powers of it. So the difference seems to be of the form $O^n-O^{m}(n>m)$ ( where $O$ is an odd number ) . So let us call the set of all such differences $\mathfrak{D}^{n}_{O}$ set of all $\left\{O^n-O^{m}\right\}$ such that the integer $m$ runs from $0$ to $n-1$ . Here For example we can start writing all such differences to see an interesting property.

Fix $O=5$. Let us take and $n=1$ and $\mathfrak{D}^{1}_{5}$ is nothing but the set of 1 element $\left\{5^1-5^0=4\right\}$ Let us now take $n=2$ and $\mathfrak{D}^{2}_{5}$ is nothing but only 2 elements $\left\{5^2-5=20,5^2-5^0=24\right\}$. Let us now take $n=3$. So the $\mathfrak{D}^{3}_{5}$ is nothing but set of 3 elements $\left\{5^3-5^2=100,5^3-5=120,5^3-5^0=124\right\}$ \( Since for $n=3$ there are only two possible $m=1,2 (3-1=2)$. Let us now take $n=4$. So the $\mathfrak{D}^{4}_{5}$ is nothing but the set of 3 elements $\left\{5^4-5^3=500,5^4-5^2=600,5^4-5=620,5^4-5^0=624\right\}$. Let us now take $n=5$. So the $\mathfrak{D}^{5}_{5}$ is nothing but the set of 4 elements $\left\{5^5-5^4=2500,565-5^3=3000,5^5-5^2=3100,5^5-5=3120,5^5-5^0=3124\right\}$.

And so on for different values of $n$. If we observe we find that the elements of the sets follow a good pattern. After trying for many such numbers I came to know the pattern. Let me explain it sir. So let us write down all such $\mathfrak{D}^{n}_{5}$ $\mathfrak{D}^{1}_{5}=\left\{4=5^0*(5-1)\right\}$ $\mathfrak{D}^{2}_{5}=\left\{20=5^1*(5-1),24=5^1*(5-1)+5^0*(5-1)\right\}$ $\mathfrak{D}^{3}_{5}=\left\{100=5^2*(5-1),120=5^2*(5-1)+5*(5-1),124=5^2*(5-1)+5^1*(5-1)+5^0(5-1)\right\}$
$\mathfrak{D}^{4}_{5} = \{500,600,620,624\}$

Here each element can be written as $\{500=5^3*(5-1),600=5^3*(5-1)+5^2*(5-1),620=5^3*(5-1)+5^2*(5-1)+5^1*(5-1),624=5^3*(5-1)+5^2*(5-1)+5^1*(5-1)+5^0*(5-1) \}.$\ Similarly $\mathfrak{D}^{5}_{5} = \{2500,3000,3100,3120 \}.$\ Here also each element can be written as $\{ 2500=5^5*(5-1),3000=5^5*(5-1)+5^2*(5-1),3100=5^5*(5-1)+5^4*(5-1)+5^3*(5-1),120=5^5*(5-1)+5^4*(5-1)+5^3*(5-1)+5^2*(5-1)+5^1*(5-1),3124=5^5*(5-1)+5^4*(5-1)+5^3*(5-1)+5^2*(5-1)+5^1*(5-1)+5^0*(5-1) \} $

Any pattern ?? .

Yes there is a pattern sir. For any $O^{n}-O^{m}$ for any odd number $O$ and any integers $n,m (m can be expressed as :

\begin{equation} O^{n}-O^{m}=\sum^{n-1}_{i=m} O^{i}*(O-1) \end{equation}

Now we can write the R.H.S of equation $(p^{x}-p^{x-z}+3)=k(l+1)$ as $p^{x}-p^{x-z}= \sum^{x-1}_{x-z} p^i.(p-1)$ After expanding the series and simplyifying we obtain $p^{x-z}(p-1).\large\frac{p^x-1}{p-1}$ $p^{x-z}(p^z-1)$

So we can equate to get $p^{x-z}(p^z-1)=k(l+1)-3.$

So we need to prove that both sides of the equation don't yield the same even number leading to a conrtradiciton. Hence I am trying that. I am here with a question again. Suppose we have an equation of this form $p^{x-z}(p^{z}-1)=k(l+1)-3$ where $p$ is an odd-prime, $z,y$ are integers and $y>0$ always ($x=2y+z, $ for some integer $x$ ) , $k$ is an odd number and $l$ is an even number.

So given such constraints and for any $p>3$ its well known that after substituting values of all variables its clear that both sides of the equation don't yield the same even number ( for any values ) . So how can we prove it ?.

I have been trying to prove it using congruences, but that didn't take me anywhere. So I wanted to ask it here.

Thank you.

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    @Iyengar There are ladies here too. I don't think it's appropriate call everybody "sir" or "Mr."2012-06-22

3 Answers 3

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Edit: this only answers the old version of the problem. The question was changed so this answer is out of date.

The original problem was $p^{x-z}(p^x-1)=\frac{n^2-1}{p^z}-3$

Suppose we have a solution to the equation. The only hypotheses we need are that $p$ be any integer $\ge 2$, and $x>z>0$.

Then $n^2=1+3p^z+p^x(p^x-1)$ $n^2=(p^x+1)^2-3(p^x-p^z)<(p^x+1)^2$ $n^2=(p^x-1)^2+p^x+3p^z>(p^x-1)^2$ So we must have $n^2=(p^x)^2$ but this would imply $n^2=0 \pmod p$, which contradicts $n^2=1 \pmod p$ from the first equation above. So there cannot be any solution.

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    Please see my answer below, if you are free sir.2012-06-21
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First, if $x=z$ then $ (p^x+3)(p^x-1)+4 = p^{2x}+2p^x+1=(p^x+1)^2 $ which is always a square.

Second, the index is wrong in the sum $ p^{x}-p^{x-z}= \sum p^i(p-1) $ it should be from $i=x-z$ to $x-1$, and after expanding the series and simplifying we get $ (p-1)p^{x-z} \sum^{z-1}_{i=0} p^i = p^{x-z}(p-1)\frac{p^z-1}{p-1} = p^{x-z}(p^z-1) $ which we could see by factoring the left side.

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    Really a very good observation, Zander. Thanks a lot sir, for your modification...2012-06-15
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Since $p^z|n^2-1$, we could write $n^2-1=k(p^z)+m$ for some m,k where $m\lt p^z$, so we could plug this into the equation. Then we would get $p^{2y}(p^z-1)=k+{m\over p^2}-3 $, since the left hand side is integer, we need the right hand side to be integer. However since $m\lt p^2$, the quantity $m\over p^2$ cannot be integer, so the equation does not hold.