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S.Iyengar
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Prove that (p^m+3)(p^a-1)+4 is not a perfect square. Here the p is an odd prime and all exponents are non-negative. Given m=2n+a, for some n>0 and for some a ( where a is an integer )
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Millennial said:You go like this.
First, expand out as follows: [itex]p^{m+a}-p^m+3p^a-3+4=p^{m+a}-p^m+3p^a+1[/itex]
Note that since p is an odd prime, this is an even number, so the only way it can be a square is that it is the square of an even number. By induction, we will show this can't hold.
First step is to disprove that this is not the square of zero, but since p is positive and we have a plus one sign, this expression is not zero. So the first step of induction is complete.
Second step is to prove that if it fails for n, it also fails for n+2. To achieve this, we acknowledge that [itex](n+2)^2=n^2+4n+4[/itex]. We will derive a contradiction from this. Since two sides must be equal, we get [itex]n^2+4n=p^{m+a}-p^m+3p^a-3[/itex]. Now, define the right hand side to be A and solve the second-degree equation to get [itex]n=-2\pm\sqrt{4-A}=-2\pm\sqrt{7-p^{m+a}+p^m-3p^a}[/itex] so we get [itex]-(n+2)^2=p^{m+a}-p^m+3p^a-7[/itex]. From here, we see that [itex]-(n+2)^2=(n+2)^2-4[/itex], so it easily follows that [itex](n+2)^2=4[/itex] and n=0. But we assumed that the expression could not hold for n=0 in the first place! Contradiction. Hence it is impossible to express this as a perfect square.
S.Iyengar said:Never mind
Curious3141 said:Please note that it's not good form to delete your original post. Deleting it after it's been replied to means that others don't know what you asked in the first place, and no one else can learn from the exchange.
Millennial said:I will be glad to answer your questions, though I strongly advise you work the proof out yourself.
1) Actually, it is sufficient that p is odd, being prime is not necessary. Just prove that the powers of an odd number are always odd, and then it will be straightforward to show.
2) You are right, I did make a mistake there, but it really does not change the course of the proof. I was testing if you would actually see my mistake. In the correct case you posted, you get an even graver contradiction. I won't post it here, you should be able to derive it from there.
Millennial said:I will be glad to answer your questions, though I strongly advise you work the proof out yourself.
1) Actually, it is sufficient that p is odd, being prime is not necessary. Just prove that the powers of an odd number are always odd, and then it will be straightforward to show.
2) You are right, I did make a mistake there, but it really does not change the course of the proof. I was testing if you would actually see my mistake. In the correct case you posted, you get an even graver contradiction. I won't post it here, you should be able to derive it from there.
Millennial said:Because we only need to prove it fails for even n, since it already fails for odd (your first question.) To do that, you start with n=0 and then use induction that covers only even numbers rather than all.
haruspex said:This is the strangest thread.
Millennial, did it not strike you as odd that you have not used any of the features of the expression A? You could substitute all sorts of expressions for A and apparently deduce the same contradiction. Clearly there was an error, which you found. But I harbour very strong doubts that you could have repaired it so easily.
S.Iyengar, I am concerned that you may have fallen into a similar error. Would you mind posting your solution?
Millennial said:I will be glad to answer your questions, though I strongly advise you work the proof out yourself.
1) Actually, it is sufficient that p is odd, being prime is not necessary. Just prove that the powers of an odd number are always odd, and then it will be straightforward to show.
2) You are right, I did make a mistake there, but it really does not change the course of the proof. I was testing if you would actually see my mistake. In the correct case you posted, you get an even graver contradiction. I won't post it here, you should be able to derive it from there.
oli4 said:Hi,
this is disproved if you take a=0 since 4 is a square
Millennial said:Yes, I think I made a mistake there. I used another logic this time and came up with a contradiction. I won't post it here, I will tell you what kind of a contradiction it is. Try to obtain it yourself.
Contradiction: Let n be the number which gives the expression you gave when squared. Then, n can't be divided by p. However, either n-4 and n+1, or n-2 and n-1 are divisable by p. From here, with some calculations, it follows that p is either 2 or irrational.
Millennial said:Yes, I think I made a mistake there. I used another logic this time and came up with a contradiction. I won't post it here, I will tell you what kind of a contradiction it is. Try to obtain it yourself.
Contradiction: Let n be the number which gives the expression you gave when squared. Then, n can't be divided by p. However, either n-4 and n+1, or n-2 and n-1 are divisable by p. From here, it is easy to see that only the first case can be valid where p is 5. However, with some effort you can show that a perfect square can be written as 5x+1 if and only if the number to be squared is in the form 5y+1. Since this can't hold with your expression (because if it did, n+1 would not be divisable by 5), we achieve a contradiction.
Millennial said:I used modular arithmetic this time, along with reductio ad absurdum.
Millennial said:I used modular arithmetic this time, along with reductio ad absurdum.
Here is the first thing I did. I used the properties of the expression you gave as follows:
[tex]n^2=p^{m+a}-p^m+3p^a+1[/tex]
[tex]n^2-1=(n-1)(n+1)=p^{m+a}-p^m+3p^a=p^a(p^m-p^{m-a}+3)[/tex]
From here, you see either n-1, n+1, or both are divisable by p (since a is nonzero.) If both were divisable, it would imply p=2, which you stated can't be true; since p is an odd prime. I hope you can carry the proof on.
Millennial said:I used modular arithmetic this time, along with reductio ad absurdum.
Here is the first thing I did. I used the properties of the expression you gave as follows:
[tex]n^2=p^{m+a}-p^m+3p^a+1[/tex]
[tex]n^2-1=(n-1)(n+1)=p^{m+a}-p^m+3p^a=p^a(p^m-p^{m-a}+3)[/tex]
From here, you see either n-1, n+1, or both are divisable by p (since a is nonzero.) If both were divisable, it would imply p=2, which you stated can't be true; since p is an odd prime. I hope you can carry the proof on.
Millennial said:That is the part which you should answer.
Hi, is it me or the original statement has changed ?S.Iyengar said:Sir, my proof has turned to be not correct. Can you give some hints in proving that ?
There "a" is non-zero sir
Millennial said:I used modular arithmetic this time, along with reductio ad absurdum.
Here is the first thing I did. I used the properties of the expression you gave as follows:
[tex]n^2=p^{m+a}-p^m+3p^a+1[/tex]
[tex]n^2-1=(n-1)(n+1)=p^{m+a}-p^m+3p^a=p^a(p^m-p^{m-a}+3)[/tex]
From here, you see either n-1, n+1, or both are divisable by p (since a is nonzero.) If both were divisable, it would imply p=2, which you stated can't be true; since p is an odd prime. I hope you can carry the proof on.
Millennial said:That is the part which you should answer.
S.Iyengar said:Sir, I have tried the following proof. Please verify whether it is right or wrong. I used t for m and [itex]\phi[/itex] for a for the sake of brevity.
So let us show that this fails in the case of [itex]n^2[/itex]. Let us start with the equation [itex] p^{t+\phi}-p^{t}+3p^{\phi}+1[/itex]. If that is a square it would be equal to some n^2 for some even [itex]n[/itex]. [itex] p^{t+\phi}-p^{t}+3p^{\phi}+1=n^2[/itex]. We prove that it will not happen based upon two cases. Let us simplify it further. [itex]p^{\phi}(p^t-p^{t-\phi}+3)= n^2-1[/itex]
\begin{equation}
p^{\phi}(p^t-p^{t-\phi}+3) = (n+1)(n-1).
\end{equation}
From the above equation its evident that [itex]p^{\phi}[/itex] may divide either [itex](n+1)[/itex] or [itex](n-1)[/itex]. So there are two cases possible.
\subsubsection{ Dividing n-1 }
As we said that [itex]p^{\phi}|(n-1) [/itex] . From here the proof turns out to be interesting. We know that [itex]n[/itex] is even, hence [itex]n-1[/itex] would be odd. So if [itex]p^{\phi}[/itex] divides [itex](n-1)[/itex] in [itex](n-1)(n+1)[/itex] it must give rise to some new term of the form [itex]d.(n+1)[/itex] where [itex]d[/itex] is odd ( Since, odd number divided by odd number would give rise to odd number ). So that [itex]d.(n+1) = (p^t-p^{t-\phi}+3) [/itex] (since from (2)). So let us prove that its a contradiction.
[itex]d.(n+1) = (p^t-p^{t-\phi}+3)[/itex]
[itex]d.(n+1) = p^t(1-p^{-\phi})+3[/itex]
[itex] d.(n+1)-3= p^t(1-p^{-\phi}) [/itex]---(3)
So here its evident that [itex]p^t | d.(n+1)-3 [/itex] . So the equation (3) can be written as
[itex]M.(d.(n+1)-3)=(1-p^{-\phi})[/itex]. Where it divides [itex]M[/itex] times ( [itex]M[/itex] can be even [itex]1[/itex] in which the same equation is obtained ).
So its clearly evident that L.H.S is an integer. So the R.H.S is
[itex] \frac{p^{\phi}-1}{p^{\phi}}[/itex]---(4)
So its evident that a prime number minus one is an even number. Prime numbers are nothing but in some sense odd numbers. So odd number minus one would be an even number. So the numerator in the (4) is a even number and denominator is the odd number. So it would never turn to an integer. Which is a contradiction.
No, you can't do that. [itex]1-p^{-\phi}[/itex] is not an integer.S.Iyengar said:[itex] d.(n+1)-3= p^t(1-p^{-\phi}) [/itex]---(3)
So here it's evident that [itex]p^t | d.(n+1)-3 [/itex] .
Norwegian said: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.
haruspex said:No, you can't do that. [itex]1-p^{-\phi}[/itex] is not an integer.
E.g. 18 = 33-32 = 33(1-3-1), but 33 does not divide 18.
Norwegian said: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.