Diophantine equation x^2=y^5+7

[Guest]

Prove that the equation has no natural solutions.
[tex]x^{2}=y^{5}+7[/tex]

[Guest]

Spoiler: show

Looking at it [tex]\bmod 4[/tex] we see that we must have [tex]y\equiv 1 \bmod 4[/tex]. Note that [tex]x^2=y^5+7[/tex] can be rewritten as
[tex]x^2+5^2 = (y+2)((y+2)(y^3-4y^2+12y-32)+80)[/tex]
The left hand side is the sum of two squares which means every prime divisor [tex]p[/tex] that is [tex]3 \bmod 4[/tex] must occur to an even power (because of the two squares theorem, a consequence of Fermat's [tex]4n+1[/tex] theorem). The two (outermost) bracketed terms on the right hand side have no prime divisors in common that are [tex]3 \bmod 4[/tex](any common divisor must divide [tex]80[/tex]). Consequently every prime that is [tex]3 \bmod 4[/tex] that divides [tex]y+2[/tex] must appear to an even power in the factorization of [tex]y+2[/tex] which is a contradiction as [tex]y+2[/tex] is [tex]3 \bmod 4[/tex].

R. Baber.

[Guest]

You can prove this older "Math Problem of the Week" problem
viewtopic.php?f=42&t=365
has no solutions in much the same way.

Hope this helps,

R. Baber.

[Guest]

... and
http://www.math10.com/forum/viewtopic.php?f=47&t=261
http://www.math10.com/forum/viewtopic.php?f=47&t=266

R. Baber.

[Guest]

...and
viewtopic.php?f=47&t=506

R. Baber.

[Math Tutor]

Next time will be a different type of problem.

[Guest]

can some help me please with this one

x^2=y^5+4

[Math Tutor]

What about it? Are there any conditions?

[Guest]

x and y are positive integers

[Guest]

x=6, y=2 is a solution (I don't think there are any other solutions for y<1000000).

Hope this helped,

R. Baber.