First we prove that for $ x\in \{-2,-1,0\}$ and less there are no roots. Case1): Assume that $ x\equiv 1 \ (mod2)$ and $ y\equiv0 \ (mod2)$ Then $ LHS\equiv 1+(-1)\equiv 0 \ (\mod2^{5})$ It's easy to see then, that we have $ y\equiv 0 \ (mod2^{3})$ required Then prove that there are no solutions in that case with infinite Fermat's regression Case2): Assume $ x\equiv y+1 \equiv 0 \ (mod2)$ Then we have $ x^{5}+32=y^{2}+1$ Now I think considering mod32 another time and Fermat's infinite regression theorem gives us no solutions in $ x,y \in N$ and considering the first line in all integers

polskimisiek wrote: Then prove that there are no solutions in that case with infinite Fermat's regression infinite Fermat's regression=infinite descent :

Yes, sorry for this, but simply I wasn't quite sure how to translate this to english

Looking $ \mod 4$, we get $ x$ odd. Now the equation is equivalent to $ (x + 2)(x^4 - 2x^3 + 4x^2 - 8x + 16) = y^2 + 1$. Now we know that $ y^2 + 1$ has no prime factor $ p \equiv 3 \mod 4$. But $ x^4 - 2x^3 + 4x^2 - 8x + 16 \equiv x^4 - 2x^3 \equiv 1 - 2 \cdot (\pm 1) \equiv 3 \mod 4$, a contradiction.

ZetaX wrote: Now the equation is equivalent to $ (x + 2)(x^4 + 2x^3 + 4x^2 + 8x + 16) = y^2 + 1$. Are you sure? Wouldn't it be more like $ (x + 2)(x^4 - 2x^3 + 4x^2 - 8x + 16) = y^2 + 1$

could anybody explain me what is the infinite Fermat's regression?

You can find the basics here: http://en.wikipedia.org/wiki/Infinite_descent

Akashnil wrote: ZetaX wrote: Now the equation is equivalent to $ (x + 2)(x^4 + 2x^3 + 4x^2 + 8x + 16) = y^2 + 1$. Are you sure? Wouldn't it be more like $ (x + 2)(x^4 - 2x^3 + 4x^2 - 8x + 16) = y^2 + 1$ Ah right, thanks. But this doesn't affect the proof. @Infinite descent: I'm sure it's not as easy as you say, so please write it out!

Well, you're quite right, it turned out that it is even not needed , but it works similar: Hmm, for the first case it should work like that: we have $ x = 2p + 1$ $ y = 8l$ Suppose we have a root: $ x_{0}^{5} + 31 = y_{0}^{2} = 64l_{0}^{2}$ There must be $ 64|LHS$ $ (2p_{0} + 1)^{5} = 64l_{0}^{2}$ $ 32p_{0}^{5} + 5*16p_{0}^{4} + 10*8p_{0}^{3} + 10*4p_{0}^{2} + 5*2p_{0} + 32 = 64l_{0}^{2}$ $ 16p_{0}^{5} + 5*8p_{0}^{4} + 10*4p_{0}^{3} +10*2p_{0}^{2}+ 5*p_{0} + 16 = 32l_{0}^{2}$ Now consider $ p_{0} = 2s_{0} + 1$ it's easy to see that than the equation does not hold Now if $ p_{0} = 2s_{0}$ we know that: $ LHS\equiv 10*2p_{0}^{2} + 5p_{0} + 16\equiv 5*16s_{0}^{2} + 10s_{0} + 16\equiv 16s_{0}^{2} + 10s_{0} + 16 \ (mod32)$ thus we know that $ s_{0} = 2s_{1}$ Now we know that $ LHS\equiv 20s_{1} + 16 \ (mod32)$ which is equivalent to: $ 5s_{1} + 4\equiv 0 \ (mod8)$ thus we aquire $ s_{1} = 2s_{2}$ $ LHS\equiv 5s_{2} + 2 \ (mod4)$ thus $ s_{2} = 2s_{3}$ We acquire $ p_{0} = 16s_{3}$ Putting that into our equation and dividing by 16: $ 16^{5}s_{3}^{5} + 5*8*16^{3}*s_{3}^{4} + 10*4*16^{2}*s_{3}^{3} + 10*2*16*s_{3}^{2} + 5s_{3} + 1 = 2l_{0}^{2}$ Now substitute $ s_{3} = 2s_{4} + 1$ (which is obvious) and divide by 2: we land in an equation: $ 16^{5}(2s_{4} + 1)^{5} + 5*8*16^{3}(2s_{4} + 1)^{4} + 10*4*16^{2}*(2s_{4} + 1)^{3} + 10*2*16*(2s_{4} + 1)^{2} + 5s_{4} + 3 = l_{0}^{2}$ Now we know that $ RHS\equiv 0\vee 1 \ (mod3)$ Now we know that: $ LHS\equiv (2s_{4} + 1)^{5} + (2s_{4} + 1)^{4} + (2s_{4} + 1)^{3} - (2s_{4} + 1)^{2} - s_{4} \ (mod3)$ Now consider three cases: I) $ s_{4}\equiv 0 \ (mod3)$ then $ LHS\equiv 1 + 1 + 1 - 1 = 2 \ (mod3)$ II)$ s_{4}\equiv 1 \ (mod3)$ then $ LHS\equiv - 1\equiv 2 \ (mod3)$ III)$ s_{4}\equiv 2 \ (mod3)$ then $ LHS\equiv 2^{5} + 2^{4} + 2^{3} - 2^{2} - 2\equiv 50 \equiv 2 \ (mod3)$ thus we acquire $ LHS\equiv 2 \ (mod3)$ which is equivalent to the fact that we have no solutions for proposedconditions The second case goes similarily but is a lot easier Well, it's quite obvious that your proof is a lot 'cleaner' than mine and requires hadly any transforms, while mine on the other hand is pretty ugly

ZetaX wrote: Akashnil wrote: ZetaX wrote: Now the equation is equivalent to $ (x + 2)(x^4 + 2x^3 + 4x^2 + 8x + 16) = y^2 + 1$. Are you sure? Wouldn't it be more like $ (x + 2)(x^4 - 2x^3 + 4x^2 - 8x + 16) = y^2 + 1$ Ah right, thanks. But this doesn't affect the proof. @Infinite descent: I'm sure it's not as easy as you say, so please write it out! how did you factorize that?

By $ a^n-b^n =(a-b)(a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1})$. Use it for $ a=x,b=-2,n=5$ here. We can even give the complete factorisation of $ x^n-y^n$ into rational irreducible polynomials: It is $ x^n-y^n = \prod_{d|n} \Phi_d(x,y)$, with $ \Phi_d(x,y)= \prod_{\substack{k=1\\\gcd(k,d)=1}}^d (x-\zeta_d^k y)$ (may not look like one, but this polynomial has integer coefficients) where $ \zeta_d=e^{\frac{2 \pi i}{d}}$.

ZetaX wrote: Now we know that $ y^2 + 1$ has no prime factor $ p \equiv 3 \mod 4$. I do not know this. Why is it?

Because think with modulo 11, $ {x}^{5} + 31 \equiv$ 10,11,12 (mod 11) and $ {x}^{2}\equiv$0,1,3,4,5,9 (mod 11) Therefore there are no integer solutions

Johan Gunardi: Because $ p|x^2 + 1$ means that $ - 1$ is a quadratic residue $ \mod p$, which isn't possible for $ p \equiv 3 \mod 4$. Alternatively, you can directly prove this fact using little Fermat: Assume $ p\equiv 3 \mod 4$ and $ p|x^2+1$, then $ 1 \equiv x^{p - 1} = (x^2)^{\frac {p - 1}{2}} = ( - 1)^{\frac {p - 1}{2}} = - 1 \mod p$, contradiction.

Heebeen, Yang wrote: Because think with modulo 11, $ {x}^{5} + 31 \equiv$ 10,11,12 (mod 11) and $ {x}^{2}\equiv$0,1,3,4,5,9 (mod 11) Therefore there are no integer solutions Maybe, You've made a mistake. The correct should be \[ {x}^{5} + 31 \equiv 8,9,10 (mod 11) \] So, your solution doesn't finish yet.

ZetaX wrote: Looking $ \mod 4$, we get $ x$ odd. Now we know that $ y^2 + 1$ has no prime factor $ p \equiv 3 \mod 4$. Thanks, but how it can be proved? Where I can read about this fact?

You would do very well if you would read my post before this one...

Oh! I'm sorry =_=... Thank you for finding my mistake

Inspired by a solution to PEN H15: $ 5 = \frac{11 - 1}{2}$, thus $ x^5 + 31 \equiv x^5 - 2 \equiv - 1, - 3 \pmod {11}$. It is enough to show that $ - 1, - 3$ are not quadratic residues mod 11. $ - 1$ is not a quadratic residue because $ 11 \neq 1 \pmod 4$. $ - 3$ is not a quadratic residue because $ \left( \frac {3}{11} \right) = 1$ by quadratic reciprocity and we showed tgat $ \left( \frac { - 1}{11} \right) = - 1$, thus $ \left( \frac { - 3}{11} \right) = - 1$.

ahmedosama wrote: No sol ... Actually taking mod 11 on both sides solves the problem.

y^2+1 divides x+2 x doesn't 1mod(4)

Hello this is mine solution (case 1) when x is even which means x=2m \forall{m}\in{Z} and also note that y is odd taking (mod 4) we know that y^{2}\equiv{1}(mod 4) and also note that (x^{5} + 31)\equiv{(x^{5}+3)}\equiv{y^{2}}\equiv{1}(mod 4) which means {x^{5}}\equiv{2}(mod 4) but note that x^{5}\equiv{32m^{5}}\equiv{0}(mod 4) and here arrives the contradiction \cdots lets do (case 2) when x is odd which means y is even now note that {(x^{5}+31)}\equiv{0}(mod 32) which means y^{2}\equiv{0}(mod 32) and {32}\mid{4m^{2}} which means m^{2} is of form 8q \forall{m,q}\in{Z} which means y^{2}\equiv{0}(mod8) now we get that y^{2}=8k=32i for some integers k,i which tells that it has no solutions by infinite descent

y^2+1 divides x+2 x doesn't 1 mod (4)

Adding $1$ to both sides we obtain $x^5 + 32 = y^2 + 1$. Now if $y$ is odd, the right-hand side is $2 \pmod{4}$, but clearly the left-hand side must be divisible by $32$, which leads to no valid solutions. Then $y$ is even, so by Fermat's Christmas Theorem, all prime factors of $x^5 + 32$ are $1 \pmod{4}$. Then since $x + 2$ is a factor, $x \equiv 3 \pmod{4}$. But plugging this in gives $x^5 + 32 \equiv 3 \pmod{4}$, but $y^2 + 1 \equiv 1 \pmod{4}$. Contradiction. Then no integer solutions exist.

$x^5 + 2^5 = y^2 + 1^2$. If $y$ odd, $2 \mid x^5 + 2^5 \implies 2 \mid x^5 \implies 32 \mid x^5 \implies 4 \mid y^2 + 1$, contradiction since $y^2 + 1\equiv 2\mod{4}$. So, $y$ even. We have $x^5 + 2^5 \equiv 1\mod{4} \implies x \equiv 1\mod{4}$. Note $x + 2 \equiv 3\mod{4}$, so it has a prime divisor $p$ such that $p \equiv 3\mod{4}$. $p \mid x + 2 \mid x^5 + 2^5 \implies p \mid y^2 + 1^2$, a contradiction by Fermat's Christmas Theorem.

Notice that $x^5+32=y^2+1$ and so from Fermat Christmas we have that every prime divisor of $x^5+32$ has to be $1 \pmod 4$, which means that $x+2 \equiv 1 \pmod 4$ so $x \equiv -1 \pmod 4$ but also $x^4-2x^3+4x^2-8x+16 \equiv 1 \pmod 4$ and replacing gives $(-1)^4-2(-1)^3+4(-1)^2-8(-1)+16 \equiv 1 \pmod 4$ and this false as LHS is $1+2+4+8+16 \equiv -1 \pmod 4$, thus no solutions can exist.