MHB Can You Find a Unique N for This Cubic and Quadratic Diophantine Equation?

[lfdahl]

Consider the equations:

$$N = x^3(3x+1) = y^2(y+1)^3$$

- where $x$ and $y$ are coprime positive integers.

Show, that there is only one possible value for $N$ and find it.

 

[Albert1]

Consider the equations:

$$N = x^3(3x+1) = y^2(y+1)^3$$

- where $x$ and $y$ are coprime positive integers.

Show, that there is only one possible value for $N$ and find it.

my solution:

Spoiler

$x^3=(y+1)^3---(1)$
$3x+1=y^2---(2)$
solution of $(1)(2),x=5,y=4,N=2000$

 

[kaliprasad]

my solution:

Spoiler

$x^3=(y+1)^3---(1)$
$3x+1=y^2---(2)$
solution of $(1)(2),x=5,y=4,N=2000$

Spoiler

x and y are co-primes that does not mean that 3x+1 and y+1 are co-primes. If they are then we need to prove it

 

[Albert1]

Spoiler

x and y are co-primes that does not mean that 3x+1 and y+1 are co-primes. If they are then we need to prove it

Spoiler

It is clear
for $1=\dfrac {(x^3)(3x+1)}{(y^2)(y+1)^3}$
if $\dfrac {x^3}{y^2}=\dfrac {b}{a}$
then $\dfrac {3x+1}{(y+1)^3}=\dfrac {a}{b}$
$a,b$ are co-primes

 

[kaliprasad]

Spoiler

It is clear
for $1=\dfrac {(x^3)(3x+1)}{(y^2)(y+1)^3}$
if $\dfrac {x^3}{y^2}=\dfrac {b}{a}$
then $\dfrac {3x+1}{(y+1)^3}=\dfrac {a}{b}$
$a,b$ are co-primes

Spoiler

Not quite correct as we can have 3x+1 = na and $(y+1)^3=nb$.

 

[lfdahl]

my solution:

Spoiler

$x^3=(y+1)^3---(1)$
$3x+1=y^2---(2)$
solution of $(1)(2),x=5,y=4,N=2000$

Spoiler

Hi, Albert!
You´re almost right. Since $gcd(x,y) = 1$, $x^3$ must divide $(y+1)^3$, i.e. $x|(y+1)$, and we conclude that: $x \le y+1$. By the same argument, we have: $y^2 \le 3x+1.$

 

[Albert1]

Spoiler

Hi, Albert!
You´re almost right. Since $gcd(x,y) = 1$, $x^3$ must divide $(y+1)^3$, i.e. $x|(y+1)$, and we conclude that: $x \le y+1$. By the same argument, we have: $y^2 \le 3x+1.$

Spoiler

ex:$x=5,y=3, gcd(5,3)=1$ but $x^3$ not divide $(y+1)^3$

 

[Albert1]

Spoiler

Not quite correct as we can have 3x+1 = na and $(y+1)^3=nb$.

Spoiler

in this case if $3x+1=na,(y+1)^3=nb$
then $x^3=nb ,y^2=na$
$x\,\,and\,\,y$ are not coprime (if $n\neq 1$)

 

[kaliprasad]

Spoiler

in this case if $3x+1=na,(y+1)^3=nb$
then $x^3=nb ,y^2=na$
$x\,\,and\,\,y$ are not coprime (if $n\neq 1$)

Spoiler

if $3x+1=na,(y+1)^3=nb$
and $x^3=b ,y^2=a$
satisfy the given condition

 

[lfdahl]

Thankyou kaliprasad and Albert for an interesting discussion and for Albert´s correct solution. I do hope we can agree on the suggested solution :)

Suggested solution:

Spoiler

Since $gcd(x,y) = 1$ it follows that $x^3|(y+1)^3$, hence $x|(y+1) \Rightarrow x \le y + 1$.
Similarly $y^2|(3x+1)$, so $y^2 \le 3x+1$. Combining the inequalities gives: $(x-1)^2 \le y^2 \le 3x+1$.
As $x$ is positive, only $x = 1,2,3,4,5$ are possible. Trying these values in $x^3(3x+1) = y^2(y+1)^3$,
we find that only $x=5, y = 4$ work, so $N = 4^2\cdot 5^3 = 2000$ is the only solution.

This is also Albert´s conclusion and final result. The solution $(5,4)$ underlines the fact, that
the factors $(3x+1) = 2^4$ and $(1+y)^3 = 5^3$ are indeed coprimes.