Disproving Statement: No Natural Numbers Satisfy y^5 + 1 = (x^7-1)/(x-1)

[Faiq]

Homework Statement

Prove that there doesn't exist natural numbers x and y such that the statement holds true.
y^5 + 1 = (x^7-1)/(x-1)

The Attempt at a Solution

I was able to simplify the term down to
y^5 / x = x^5 + (x^5-1)/(x-1)
Not sure what to do with it

 

[member 587159]

Homework Statement

Prove that there doesn't exist natural numbers x and y such that the statement holds true.
y^5 + 1 = (x^7-1)/(x-1)

The Attempt at a Solution

I was able to simplify the term down to
y^5 / x = x^5 + (x^5-1)/(x-1)
Not sure what to do with it

I'm not sure but maybe you can use this:

Note that $\frac{x^7 - 1}{x - 1} = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$.

 

[Faiq]

Yes in that form if x is not equal to 3k -1 it is true but if it is its not coming true

 

[Faiq]

y5 = x(x+1)(x4+x2+1)
gcd(x,x+1) = gcd(x,x4+x2+1 )=1
however
gcd(x+1,x4+x2+1)=gcd(x+1,3)

 

[Faiq]

How should I proceed?

 

[haruspex]

How should I proceed?

Consider a prime factor p of x. Suppose it divides x exactly k times. What can you deduce about y in relation to p?

 

[haruspex]

@Faiq , have you lost interest in this? My method works.

 

[Delta2]

@Faiq , have you lost interest in this? My method works.

Don't know about faiq but I am interested in this. All I can infer from your hint is that y has also all the prime divisors of x at least once. After that what follows?

 

[haruspex]

at least once.

Can you be more exact?

 

[Delta2]

Can you be more exact?

Nope, I don't think I can prove that y has all the prime divisors of x exactly the same times that is essensially that y is a multiple of x? is this what you trying to tell me?

 

[haruspex]

essensially that y is a multiple of x?

Almost.
If p divides x exactly k times, how often does it divide y⁵? What does that tell you about k?

 

[Delta2]

It divides $y^5$ at least k times but I got no clue what does this tells me about $k$?

 

[haruspex]

It divides $y^5$ at least k times but I got no clue what does this tells me about $k$?

Can it divide y⁵ more than k times?

 

[Delta2]

Can it divide y⁵ more than k times?

I guess not if I take into account some of the posts of @Faiq about the gcd..

Sorry but anyway I don't see where you trying to lead me, but I just want to know what theorems/lemmas of number theory you are using. Do you use only the fundamental theorem of number theory along perhaps with some of its immediate corollaries (would be good to mention which ones you use though).

 

[haruspex]

I guess not if I take into account some of the posts

There are two factors each side: (y⁵)(x-1)=(x)(x⁶-1). If p divides x k times but y⁵ more than k times, p needs to divide the other factor on the right. Is that possible?
So what can we say about the relationship between x and y⁵? What about the other factors of y⁵?

I am not using any theorems or lemmas, just simple logic.

 

[Delta2]

There are two factors each side: (y⁵)(x-1)=(x)(x⁶-1). If p divides x k times but y⁵ more than k times, p needs to divide the other factor on the right. Is that possible?
So what can we say about the relationship between x and y⁵? What about the other factors of y⁵?

I am not using any theorems or lemmas, just simple logic.

Seems to me you are using some lemmas besides the fundamental theorem of number theory but for you must be so obvious and simple facts that you call them simple logic. However for me they are not so obvious. For example you seem to use a little lemma that "if a prime p divides a product ab then p divides a or p divides b" which is a corollary from the fundamental theorem.

(I wonder why it seems so obvious to you that p cannot divide $x^6-1$)
ok anyway seems to me you implying that $y^5=cx$ where c contains the other prime factors of y^5. But still can't see where do we go from that. c seems complete mystery to me I can't make anything about it. We want to prove that y cannot be integer if x is integer right?

 

[haruspex]

(I wonder why it seems so obvious to you that p cannot divide $x^6-1$)
ok anyway seems to me you implying that $y^5=cx$ where c contains the other prime factors of y^5.

The multiples of p are at intervals of p, so two consecutive numbers cannot have a factor in common.

Since c and x are coprime, and their product is a fifth power, what does that tell you about c and x individually?

 

[Delta2]

The multiples of p are at intervals of p, so two consecutive numbers cannot have a factor in common.

That's another little lemma that you classify as simple logic I guess (hehe), if x and x+1 have a common factor then their difference which is 1 must be a multiple of their common factor, ok got it.

Since c and x are coprime, and their product is a fifth power, what does that tell you about c and x individually?

Here is the critical piece of the proof for me, I couldn't even imagine or notice that c and x are coprime...If I understand correctly since their product is a fifth power it means that they aren't coprime afterall which is a contradiction correct?

 

[haruspex]

c and x are coprime

Right.

since their product is a fifth power it means that they aren't coprime after all

Wrong.

 

[Delta2]

Hmmm ok it means that they are both a fifth power. But if x is a fifth power then c (which is $(x+1)(x^4+x^2+1)$) cannot be a fifth power?

 

[haruspex]

it means that they are both a fifth power

Right.

c (which is $(x+1)(x^4+x^2+1))$ cannot be a fifth power?

Not so fast.. this is the clever bit.
Cancelling the x factor and writing c=d⁵, we have: d⁵=x⁵+x⁴+x³+x²+x+1.
Can you spot the clincher?

 

[Delta2]

Well not sure, apparently it is $d>x$ so if we set $d=x+e$ e>0 integer then $(x+e)^5=x^5+x^4+x^3+x^2+1$ am I on the right track?

 

[haruspex]

Well not sure, apparently it is $d>x$ so if we set $d=x+e$ e>0 integer then $(x+e)^5=x^5+x^4+x^3+x^2+1$ am I on the right track?

Yes. How small can e be?

 

[Delta2]

Yes. How small can e be?

Well if I do the binomial expansion of $(x+e)^5$ and since x and e are positive, seems to me there are no positive integer values of e for which this equality can be true. The smallest value for e is 1 but still it doesn't work . Is that all?

 

[haruspex]

Well if I do the binomial expansion of $(x+e)^5$ and since x and e are positive, seems to me there are no positive integer values of e for which this equality can be true. The smallest value for e is 1 but still it doesn't work . Is that all?

That's it.

 

[Faiq]

A

 

[Faiq]

Here's a small solution on my part
y^5 = x(x5+x4+x3+x2+x+1)
gcd(x,x5+x4+x3+x2+x+1) = 1
These two terms are coprime

Case 1: X is a fifth power but C is not
y5=cx c = x5+x4+x3+x2+x+1
y5 = c(p1*p2*p3...pn)^5
k^5(p1*p2*p3...pn)^5= c(p1*p2*p3...pn)^5
k^5 =c
C is also a fifth power
Contradiction to our base case
Same contradiction comes when C is considered a fifth pwer but X is not

Case 2: X and C are both a fifth power
c = d5
d5 = x5+x4+x3+x2+x+1
(x+e)^5 = x5+x4+x3+x2+x+1
No positive value of e exists for which the equality is true (Although I am unsure on this part)

Case3: X and C both arent a fifth power
Not sure how to do this one

 

[Delta2]

Here's a small solution on my partHowever I cannot understand the argument to prove that two coprime numbers cannot make a fifth power

I thought so too, but @haruspex corrected me, the only way for 2 coprime numbers to make a fifth power is if they are both a fifth power of some numbers (not necessarily the same). because then $x=a^5, c=b^5, y^5=cx=(ab)^5$,So $x$ and $c=x^5+...+x+1$ have to be fifth powers and then you got to follow posts #21 to #25 to see what happens.

$(x+e)^5$ cannot be equal to $x^5+...+x+1$ IF x and e are positive integers. To see that, you have to do binomial expansion of $(x+e)^5$, gather all terms of the equality $(x+e)^5-x^5-...-x-1=0$ in one side and then you ''ll have that a sum of (positive) terms is equal to zero which cannot be. The terms are positive because x is positive and because the smallest value e can have is 1.

 

[Faiq]

Case3: X and C both arent a fifth power
X = p1.p2.p3.p4...pn
C = q1.q2.q3.q4...qn
y5 = (p1.p2.p3.p4...pn)(q1.q2.q3.q4...qn)
Since the primes factor are not in the order of five this cannot be true too

Is this type of an argument valid?

 

[Faiq]

I thought so too, but @haruspex corrected me, the only way for 2 coprime numbers to make a fifth power is if they are both a fifth power of some numbers (not necessarily the same). because then $x=a^5, c=b^5, y^5=cx=(ab)^5$,So $x$ and $c=x^5+...+x+1$ have to be fifth powers and then you got to follow posts #21 to #25 to see what happens.

$(x+e)^5$ cannot be equal to $x^5+...+x+1$ IF x and e are positive integers. To see that, you have to do binomial expansion of $(x+e)^5$, gather all terms of the equality $(x+e)^5-x^5-...-x-1=0$ in one side and then you ''ll have that a sum of (positive) terms is equal to zero which cannot be. The terms are positive because x is positive and because the smallest value e can have is 1.

I don't think its necessary for both to be a fifth power consider this
y5 = (ab)5 = a5.b5
C = a5.b
x = b4
y5 = cx = (a5.b)(b4) = a5.b5 =(ab)5

 

[Delta2]

I don't think its necessary for both to be a fifth power consider this
y5 = (ab)5 = a5.b5
C = a5.b
x = b4
y5 = cx = (a5.b)(b4) = a5.b5 =(ab)5

But then C and x wouldn't be coprime they would have common factor b.

 

[Faiq]

Oh yeah right thanks

 

[Delta2]

Btw you are right about case 1 and 3 we just omitted them because they kind of intuitively obvious that they lead to contradiction.

 

[Faiq]

Thank you very much for the help