This is a question French mathematician Pierre de Fermat posed to the English mathematicians of his time. "Prove that x=5 and y=3 are the only positive integer values for which $x^2 +2=y^3$." I have been trying to come up with an answer for days using modular arithmetic to no avail. Any such method would therefore be appreciated as well as other insights.
123390
$x^2 + 1 = z = y^3 - 1$  Why $z = 26 $ and only $26$ ? Is there an elementary proof of that ?
38113
The problem from Burton: show that the equation $n^2 + (n+1)^2 = m^3$ has no solution in the positive integers.  So far, I can see that gcd($n$,$n+1$)$=1$ and $m \equiv_4 1$ and $m=a^2 + b^2$ for some integers a,b. I'm guessing I need to reach a contradiction.  At this point, I am stuck. Any hints?
44142
I'm not sure if this proof is valid or not and could use some feedback.  Let P be the supposition that $2^n \le 2^{n+1}-2^{n-1} - 1$ for all positive integers n.   Base Case: Let n=1  $2^1 \le 2^2-2^0-1$  $2\le2$   Suppose that P holds true for some positive integer n=k  $$2^{k+1}=2*2^k$$  By inductive hypothesis:  $$2*2^k \le 2(2^{k+1}-2^{k-1}-1)$$  $$2^{k+1} \le 2^{k+2}-2^{k}-2$$   Since we know $2^{k+1} \le 2^{k+2}-2^{k}-2$, we also know $2^{k+1} \le 2^{k+2}-2^{k}-1$ because the former is always smaller.  This proves that: $$2^k \le 2^{k+1}-2^{k-1} - 1\Rightarrow 2^{k+1} \le2^{k+2}-2^{k}-1 $$
24008
just so this doesn't get deleted, I want to make it clear that I already know how to solve this using the UFD $\mathbb{Z}[\sqrt{-2}]$, and am in search for the infinite descent proof that Fermat claimed to have found.  I've alaways been fascinated by this Diophantine equation $x^2+2=y^3$ in particular ever since I saw it, and I still have no clue how to attack it without $\mathbb{Z}[\sqrt{-2}]$. What's disappointing is that no one else seems interested in the hunt (an elementary proof using infinite descent). I know it's been studied extensively, and there have even been generalizations, such as Mordell's equation. However, I've never seen Fermat's original proof that $(x,y)=(\pm 5, 3)$ is the only integer solution. Obviously, Fermat probably knew nothing of UFD's, which is why I believe there has to be an infinite descent proof like he claimed. Has anyone apart from him actually seen this proof? People mention it all the time, yet I can't find anything about it. As I said, I know that it involved infinite descent, but I've never seen it anywhere and no one seems to have any idea about it.  Does anyone have ideas for this approach? I mean, infinite descent seems more effective for showing a contradiction, e.g. showing there are no solutions. But how could it work here? Also, why isn't it published anywhere in all this time? Could it really be that only Fermat knew his method of descent well-enough to make this problem submit to it?  Thanks!
52080
May $y=e^x$ be satisfied with both $x$ and $y$ be positive integers?  I think it is not possible as $e$ ,a transcendental number, when multiplied by itself would never result in rational number.  Am I right?
7574
Find all integer solutions to $x^2+4=y^3$.  Some obvious solutions are $(x,y)=(\pm2,2)$. Are these the only ones?
109607
Consider the set of positive integers $n \in \mathbb {Z}&gt;0$ such that $\dfrac{n}{2}$ is a perfect square, $\dfrac{n}{3}$ is a perfect cube, and $\dfrac{n}{5}$ is a perfect fifth power; that is to say, such that there exist integers $a, b, c \in \mathbb{Z}$ such that $\dfrac{n}{2} = a^2$, $\dfrac{n}{3} = b^3$, $\dfrac{n}{5} = c^5$.  Let $n_0$ denote the smallest element in this set (it exists by the well-ordering principle). Determine the numerical value of $n_0$.
35948
Find all solutions of the equality $y^2=x^3+23$ for integers $x,y$   I guess, that x cannot be even. because we can apply mod 4 test  say $x=2k$ then $y^2=8k^3+23\equiv3\mod(4)$   but, this is not possible, since the square of an integer is congruent to $0$ or $1$ $\mod (4)$  so $x$ is of the form $4k+3$ or $4k+1$  $x=4k+3$ is also not possible for the same reason but for the last case the test fails.  any hints ?
10800
There is a 'nice' relationship between the integers (1,5) and (2,3) as  $$1+5=2 \times 3;$$ $$1\times 5 = 2 + 3.$$  So I tried to find all positive integers pairs $(a, b)$ and $(c, d)$ such that $$a+b=c \times d;$$  $$a\times b = c + d.$$  To find this, $a, b, c, d$ must satisfy $$(a+1)(b+1)=(c+1)(d+1).$$ However, this condition is only necessary but not sufficient. Any idea?
61045
This is Problem 6 of the 2007 Indian National Math Olympiad (INMO).  If $x, y, z$ are positive real numbers, prove that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2).$  My best idea was to expand this and simplify. Although that doesn't look very feasible. Another idea is to see that $x^2+y^2+xy \geq x^2+y^2$. Then we just have to show that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(x^2+y^2)(x^2+z^2)(y^2+z^2)$ if that is even true.
102108
Could anybody help me some ideas on the below problem:  Let $n$ be a positive integer. Prove that there are infinite pairs of positive integer $(x\, y)$ such that $$x^2+x+1=(y^2+y+1)(n^2+n+1).$$ Thanks in advance.
76358
How to solve the Diophantine equation $x^2 + 2 = y^3$ with $x,y&gt;0$ ? ($x,y$ are integers.)
93740
I need to find for which values of $a\in\mathbb{N}$ the following happens:  $$a-2\ |\ a^3+4$$  that is, for which values of $k\in\mathbb{Z}$ the following holds:  $$a^3+4 = k(a-2)$$  I could not find anything useful from that.  Usually for a proof involving division, for example, proving $8|3^{2n}+7$ I'd suppose $3^{2k}+7 = 8k$ and prove by induction that it also holds for $k+1$, but for this case is diferent.
68551
Use mathematical induction to show that:  $1^3 + 3^3 + 5^3 + ... + (2n + 1)^3 = (n+1)^2(2n^2+4n+1)$ whenever $n$ is a positive integer.  I am able to solve this problem, but the thing that confuses me on this is the Basis Step.  P(1): $(2(1) + 1)^3$ $\neq$ $4(7)$, does this mean that this problem can not be proved using mathematical induction, or am I missing something? Thanks.
53577
We now that Fermat's last theorem is true so there are not positive integer solutions to $$x^n+y^n=z^n$$ for $n\in\mathbb{N}$ and $n&gt;2$.  But what about if $n\in\mathbb{R}$ or $n\in\mathbb{R}^+$?
104049
Prove that there does not exist an integer solution for the diophantine equation $x^5 - y^2 = 4$.   It's obvious that $x$ and $y$ are of the same parity. We can also claim that if $x$ is odd, then it is $1 \pmod 4$. Also, if $x$ and $y$ are even, then $y \equiv 2 \pmod 4\text{ since } x^5 \text{ is a multiple of } 32$ and if $y$ were a multiple of $4$ then it would be at a distance of at least $16$ from $x^5$ or be equal to it. These are my observations.  How should I proceed with the proof? Please give hints.
11458
Prove, by contradiction, that, if $n$ is a positive integer such that $n^3 − n − 6 = 0$, then, for every positive integer $m$ with $m \not = n$, $m^3  − m − 6 \not = 0$.    Proposition: If $n$ is a positive integer such that $n^3 − n − 6 = 0$, then, for every positive integer $m$ with $m \not = n$, $m^3  − m − 6 \not = 0$.  Hypothesis: $n$ is a positive integer such that $n^3 − n − 6 = 0$.  Conclusion: For every positive integer $m$ with $m \not = n$, $m^3  − m − 6 \not = 0$.    My workings  A (Hypothesis): $n$ is a positive integer such that $n^3 − n − 6 = 0$.  A1 ($\neg B$): There exists a positive integer $m$ with $m \not = n$ such that $m^3  − m − 6 = 0$.  A2: $n^3 - n = 6$  $\implies n(n^2 - 1) = 6$ where $n \in \mathbb{Z}^+$  A3: $m^3 - m = 6$  $\implies m(m^2 - 1) = 6$ where $m \in \mathbb{Z}^+$ and $m \not = n$.  A4: $n(n^2 - 1) = m(m^2 - 1)$ where $m \not = n$.  $\implies \dfrac{n(n^2 - 1)}{m(m^2 - 1)} = 1$ where $m \not = n$.  Since $m,n \in \mathbb{Z}^+$, $(n^2 - 1) \ge 0$ and $(m^2 - 1) \ge 0$. Therefore, $n(n^2 - 1) = m(m^2 - 1)$ iff $n = m$.   Contradiction.  $Q.E.D.$    I would greatly appreciate it if people could please take the time to provide feedback on the correctness of my proof. Given that $m \not = n$, did I construct my proof correctly? Or is it incorrect to structure it this way if $m \not = n$?