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

[Math Tutor]

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

[Guest]

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

[Guest]

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

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

[Guest]

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

[Divyas]

Hi TEACHER,

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

[Guest]

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

[Guest]

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

[Guest]

Исходное уравнение можно решить ещё другим способом.
[tex]x^{2 }+5= y^{3 }[/tex]
[tex]x=t+1[/tex]
[tex][t^{2 }+2t+6]- y^{3 }=0[/tex], [tex]t=2m+1[/tex]
[tex]x^{2 }+6= y^{3 }+1[/tex], [tex]y+1=2(2k+1)[/tex]

[tex][( t^{2 }-1)-21]={[(y+1)-4][y^{2 }+3y+9]}[/tex]

[Guest]

Как здесь редактировать текст? (Исправляю опечатку.)
Исходное уравнение можно решить ещё другим способом.
\displaystyle x^{2 }+5= y^{3 }x
2
+5=y
3

\displaystyle x=t+1x=t+1
\displaystyle [t^{2 }+2t+6]- y^{3 }=0[t
2
+2t+6]−y
3
=0, \displaystyle t=2m+1t=2m+1
\displaystyle x^{2 }+6= y^{3 }+1x
2
+6=y
3
+1, \displaystyle y+1=2(2k+1)y+1=2(2k+1)

\displaystyle [( t^{2 }-1)-20]={[(y+1)-4][y^{2 }+3y+9]}[(t
2
−1)−20]=[(y+1)−4][y
2
+3y+9]