Prove that the equation x^2+5=y^3 has no solutions

by **Math Tutor** » Tue Feb 15, 2011 5:15 pm

Prove that the equation $x^2+5=y^3$
has no solutions for $x, y \in N$

by **Guest** » Sat Jul 09, 2011 12:25 pm

Rewrite the equation as $x^2+4 = y^3-1 = (y-1)((y-1)^2+3y)$ .
If a prime number $p$ divide $x^2+4$ then either $p=2$ or $p\equiv 1\pmod{4}$ . This implies that there is no number $d\equiv 3\pmod {4}$ that divide $x^2+4$ .
<-> If $y$ is odd,
Then $(y-1)^2+3y\equiv 1\pmod{4}$ which leads to $y-1\equiv 2 \pmod{4}$ which is impossible.
<-> If y is even,
then $y\equiv 2\pmod{4}$ wich implies that $(y-1)^2+3y \equiv 3\pmod{4}$ which is again impossible.
Thus the equation deosn't have any solution.

by **Guest** » Sat Jul 09, 2011 4:31 pm

I read the solution but I could not understand why this is true:

If a prime number $p$ divide $x^2+4$ then either $p=2$ or $p\equiv 1\pmod{4}$

by **Guest** » Sat Jul 09, 2011 5:15 pm

It is a well known fact, I wonder whether you know about the legender symbole, actually, if an odd prime number $p$ divide $x^2+4$ then $-1$ is a quadratic residue modulo $p$ which implies by Gauss's Lemma, that $p\equiv 1\pmod{4}$ . If you don't know about the legendre symbole, you can prove that fact using only Fermat's Theorem. In fact, if $p$ is an odd prime and $p|x^2+4$ then $(-4)^{\frac{p-1}{2}} \equiv x^{p-1}\pmod{p}$ this implies after Fermat liltle theorem that $(-1)^{\frac{p-1}{2}}=1$ or again $p\equiv 1\pmod{4}$

by **Divyas** » Fri Jul 22, 2011 5:33 am

Hi TEACHER,

It's not possible for me to prove this equation it looks simple any ways is that comes in maths olympiad also.

by **Guest** » Fri Jul 22, 2011 7:55 am

Another fact: if $p$ is prime of the form $4k+3$ and $p$ divides $a^2+b^2$ then $p$ divides $a$ (hence $p$ divides $b$ ). Prove it!
So if $p$ divides $x^2+2^2$ and $p$ is of the form $4k+3$ , then $p$ divides $2$ . Contradiction!

by **Guest** » Wed Jul 18, 2012 3:30 am

See x must be even then y is odd.x=2k,y=2l+1,we get $2(k^2+1)=l(4l^2+6l+3)$ ,Again l even l=2x we get
$k^2+1=x(16x^2+12x+3)$ ;
lest hand side has no prime divisor of the form 4k+3 from quadratic residue but in RHS $16x^2+12x+3$ is of form 4k+3 so it must have a prime divisor of form 4k+3 which is a contradiction.So no solution.

Prove that the equation x^2+5=y^3 has no solutions

Prove that the equation x^2+5=y^3 has no solutions

Re: Prove that the equation x^2+5=y^3 has no solutions

Re: Prove that the equation x^2+5=y^3 has no solutions

Re: Prove that the equation x^2+5=y^3 has no solutions

Re: Prove that the equation x^2+5=y^3 has no solutions

Re: Prove that the equation x^2+5=y^3 has no solutions

Re: Prove that the equation x^2+5=y^3 has no solutions

Who is online