Integer solution

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 Joined: Sat Jan 23, 2016 7:55 pm
Integer solution
Prove that ,y^2=x^3+7 has no integer solution...
 nahin munkar
 Posts: 81
 Joined: Mon Aug 17, 2015 6:51 pm
 Location: banasree,dhaka
Re: Integer solution
SOLUTION 1:
From the equation, we can easily see, x & y are of different parity. So,
Case 1 :At first,we let, x = 2p and y=2q+1
then, from equation, $4q^2$+4q +1 = $8p^3$ + 7 => $4(q^2 +q)$ = 4($ 2p^3$ +1) +2 .Here, 4 divides LHS but not RHS.Contradiction!
Case 2:Here, x=2p+1 ,y=2q as 2nd case. We will find contradiction here also like 1st case. Try to find it.I left it to you. Hence,[Proved]
From the equation, we can easily see, x & y are of different parity. So,
Case 1 :At first,we let, x = 2p and y=2q+1
then, from equation, $4q^2$+4q +1 = $8p^3$ + 7 => $4(q^2 +q)$ = 4($ 2p^3$ +1) +2 .Here, 4 divides LHS but not RHS.Contradiction!
Case 2:Here, x=2p+1 ,y=2q as 2nd case. We will find contradiction here also like 1st case. Try to find it.I left it to you. Hence,[Proved]
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss
 nahin munkar
 Posts: 81
 Joined: Mon Aug 17, 2015 6:51 pm
 Location: banasree,dhaka
Re: Integer solution
SOLUTION 2:
OK.We take 3 cases.
case1. Let, x=y. Then it can be easily told that no solution.
case2.Let, x>y. if x,y positive,then,$x^3$>$y^3$>$y^2$. so,$x^3$>$y^2$. & $x^3$$y^2$>0.....(1). we get from equation, $x^3$$y^2$=7.contradiction. Now,if x,y negative(y may be positive),then here,$x^3$<$y^2$(as $y^2$ is always positive but $x^3$ is negative).then from equation,L.H.S. is positive & in R.H.S. $x^3$ is negative adding with 7 equals to L.H.S. which is positive that can easily be checked that no solution.
case3. Let, x<y.
Now, let $y=a(mod7)=>y^2=a^2(mod7)$ .let,$x=b(mod7)=>x^3=b^3(mod7)$.now,$a>0$,$b>0$ and $a<y$ ,$b<x$
Then, $y^2x^3$=$[7(r)+a^2][7(p)+b^3]$>$a^2b^3>0$
Now,
7$y^2x^3$
=>7$a^2b^3$
Now, $a^2b^3=7$ [infinite descent!!!]
Here,Infinite descent occurs.So, $contradiction!$
SO,There is no solution in integer. [Proved]
OK.We take 3 cases.
case1. Let, x=y. Then it can be easily told that no solution.
case2.Let, x>y. if x,y positive,then,$x^3$>$y^3$>$y^2$. so,$x^3$>$y^2$. & $x^3$$y^2$>0.....(1). we get from equation, $x^3$$y^2$=7.contradiction. Now,if x,y negative(y may be positive),then here,$x^3$<$y^2$(as $y^2$ is always positive but $x^3$ is negative).then from equation,L.H.S. is positive & in R.H.S. $x^3$ is negative adding with 7 equals to L.H.S. which is positive that can easily be checked that no solution.
case3. Let, x<y.
Now, let $y=a(mod7)=>y^2=a^2(mod7)$ .let,$x=b(mod7)=>x^3=b^3(mod7)$.now,$a>0$,$b>0$ and $a<y$ ,$b<x$
Then, $y^2x^3$=$[7(r)+a^2][7(p)+b^3]$>$a^2b^3>0$
Now,
7$y^2x^3$
=>7$a^2b^3$
Now, $a^2b^3=7$ [infinite descent!!!]
Here,Infinite descent occurs.So, $contradiction!$
SO,There is no solution in integer. [Proved]
Last edited by nahin munkar on Wed Apr 27, 2016 4:13 pm, edited 4 times in total.
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss

 Posts: 12
 Joined: Sat Jan 23, 2016 7:55 pm
Re: Integer solution
Thanks for helping...
 nahin munkar
 Posts: 81
 Joined: Mon Aug 17, 2015 6:51 pm
 Location: banasree,dhaka
Re: Integer solution
You are most welcomeMd Ashraful Kader wrote:Thanks for helping...
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss
 nahin munkar
 Posts: 81
 Joined: Mon Aug 17, 2015 6:51 pm
 Location: banasree,dhaka
Re: Integer solution
Here,I get another solvent solution:
\(y^2 + 1 = x^3 + 8\)(adding 1 both sides).
now we have \(y^2 + 1 =(x+2)(x^22x+4)\).
Here, a prime divisor of LHS must be of the form \(4k+1\)[fermat sum of square].
So, both factors of RHS are \(\equiv 1 \pmod 4\). Thus \(x+2\equiv 1 \pmod 4\). so,we get \(x\equiv 1 \pmod 4\), now in the other factor we have \((1)^2 +2 + 4\) which is \(\equiv 3 \pmod 4\) so the LHS is not \(\equiv 1 \pmod 4\) and as it's not possible,there is no solution exists.
\(y^2 + 1 = x^3 + 8\)(adding 1 both sides).
now we have \(y^2 + 1 =(x+2)(x^22x+4)\).
Here, a prime divisor of LHS must be of the form \(4k+1\)[fermat sum of square].
So, both factors of RHS are \(\equiv 1 \pmod 4\). Thus \(x+2\equiv 1 \pmod 4\). so,we get \(x\equiv 1 \pmod 4\), now in the other factor we have \((1)^2 +2 + 4\) which is \(\equiv 3 \pmod 4\) so the LHS is not \(\equiv 1 \pmod 4\) and as it's not possible,there is no solution exists.
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss