I'm trying to prove/disprove that the equation  $2^x+2^y=2^z$  where $x$, $y$ and $z$ are positive integers, only have trivial solutions. The obvious case where this is true would be for $x=y$ but I'm not certain if any solutions exits for $x\ne y$  It feels like there are no other solutions but I'm not sure how one would formally prove it. Anyone have any tips on how to do this?   Sorry if this is a trivial question / known problem.
122129
If we are given $j,k \geq 0, j&gt; k$ and $j,k$ are integers, can $2^j + 2^k$ be ever expressed as $2^n$ where $n \geq 0$ and is an integer?  What I said:     Suppose it can. Then for some $0 \leq n \in \mathbb{N}: 2^j + 2^k = 2^n$. We know that $j&gt;k$ so $j=k+s$ for some $0 &lt; s \in \mathbb{N}$. So: $2^j + 2^k = 2^{k+s} + 2^k = 2^k \cdot 2^s + 2^k = 2^k \cdot (2^s + 1)$. We see that $2^s + 1$ can be an even number only if $s=0$ but we know that $s &gt; 0$, thus we have that $2^j + 2^k$ which is an even number is equal to an odd number (even * odd), thus it can't be.   Is that correct?
105936
Prove that there are no positive integers $x$ and $y$ such that $x^3 + y^3 = 10^3$.   This is a homework question, and I understand that its part of Fermat's Last Theorem, but when I looked that up to try to figure out the homework, I realized that it is way further than what we've learned in class so far. We have to prove this, but the only way I can think to do it is by exhaustion, which would be extremely lengthy. Any little shove in the right direction would be appreciated. Thanks!
107383
How can we find positive integers solutions $(x,y,z)$, where $\gcd(x,y,z)=1$ for the equation: $$1/x^2+1/y^2=1/z^2$$  Can we conclude that $x$ and $y$ are not coprimes for it to have solution?
48365
show that: For any postive ineteger $n$,then the equation    $$n=x^2+y^2+z^3$$   always have integer solution   My idea: such as $n=1$,then we have $$1=0^2+0^2+1^3$$ $$2=0^2+1^2+1^3$$ $$3=1^2+1^2+1^3$$ $$4=2^2+0^2+0^3$$ $$5=1^2+2^2+0^3$$ $$6=1^2+2^2+1^3$$ $$7=2^2+2^2+(-1)^3$$ $$8=0^2+0^2+2^3$$ $$9=1^2+0^2+2^3$$ $$10=1^2+1^2+2^3$$ $\cdots\cdots\cdots$  But for  general $n$, How prove it?
103821
I saw this problem recently and found an elegant solution to it, and was curious to see if anybody would think of something else. Nice solutions to nice problems are fun to see!  Problem: Prove that, for all non-negative integers $k$, there exist integers $a$, $b$ such that $a^2+b^2=5^k$. Bonus: Prove that there exist such $a$ and $b$ that furthermore satisfy $(a,b)=1$.
75390
Recently I asked a question regarding the diophantine equation $x^2+y^2=z^n$ for $x, y, z, n \in \mathbb{N}$, which to my surprise was answered with the help complex numbers. I find it fascinating that for a question which only concerns integers, and whose answers can only be integers, such an elegant solution comes from the seemingly unrelated complex numbers - looking only at the question and solution one would never suspect that complex numbers were lurking behind the curtain!  Can anyone give some more examples where a problem which seems to deal entirely with real numbers can be solved using complex numbers behind the scenes? One other example which springs to mind for me is solving a homogeneous second order differential equation whose coefficients form a quadratic with complex roots, which in some cases gives real solutions for real coefficients but requires complex arithmetic to calculate.  (If anyone is interested, the original question I asked can be found here: )  EDIT:  I just wanted to thank everyone for all the great answers! I'm working my way through all of them, although some are beyond me for now!
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My number theory is terrible so I don't know what "class" of problem this secretly is.  I'm looking for all positive integer solutions to the equation:     $M^2=5N^2+2N+1$   That is, I want positive integer $M$ and $N$ to make the above true.  I've got the obvious solution ($N=0$, $M=1$) but I don't know how to go about getting more solutions.  It has been suggested to me that there should be infinitely many solutions, and I would like to find them all.  I could transform it to look like Pell's equation by completing the square on the right, but it won't have integer coefficients (or you could multiply it through by the denominators, but then it wouldn't look like Pell's equation), so I don't think that helps much.  I don't know enough number theory to guess at other things, but I'm happy to read something on this topic.
67668
I've been struggling with this all day today. I imagine it's not very hard, but my algebra skills are terrible. So, how can I show that if $m&gt;n$ and $a$ is a positive integer, then $$a^{2^n}+1 \mid a^{2^m}-1.$$ I just can't get a coherent picture of what I'm supposed to do. If this too (see yesterday's question) boils down to the Unique Factorization Theorem, I would be very grateful for any intuition on how to think about this!
31440
$x, y, z \in \mathbb{N}$,    $\gcd(x, y) = 1$  prove that $x^4 + y^4 = z^2$ has no solutions.  It is true even without $\gcd(x, y) = 1$, but it is easy to see that  $\gcd(x, y)$ must be $1$
65035
I need help proving the following statement:  For all integers $x$ and $y$, if $x^3+x = y^3+y$ then $x = y$  The statement is true, I just need to know the thought process, or a lead in the right direction. I think I might have to use a contradiction, but I don't know where to begin.  Any help would be much appreciated.
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 $n\in\mathbb{R}^+$?
31064
As far as I know, we are searching solutions in set of positive integers for $x^n + y^n = z^n$  for $n &gt; 2$. There are many proofs are stated that Fermat is true and there is no solutions for this equations. Now my question is, can we have solutions in complex numbers field? If yes, how to find them.   Thanks in advance.
59246
Find all positive integers x, y, z which satisfy the equation $(x^2 + 1)(y^2 + 1) = z^2 + 1$ given that $(x^2 + 1)$ and $(y^2 + 1)$ are both primes.  It seems trivial that the only set of integers x, y and z that work are $(1^2 + 1)(2^2 + 1) = 3^2 + 1$, which is equivalent to $2 * 5 = 10$, but how would I go about proving this, or are there any other sets for which the following equation works out?  Thanks in advance.
47525
Find all integer solutions of the equation $a^2+ b^2 + c^2=a^2 b^2$.  This is one of the questions we presented in one session to contest preparation PUTNAM. It turns out that I can't get from the problem. Could someone just give me a hint? (Please, don't give me the answer. Simply, an argument that can help me advance in the problem or theorem might suffice.)
15604
1) $D$  is a positive integer, find all rational solutions of Pell's equation  $$x^2-Dy^2=1$$  2) What about $D\in\Bbb Q$ ?
92099
Find all real solutions for $x$ in $2(2^x−1)x^2+(2^{x^2}−2)x=2^{x+1}−2$.   I started off by dividing $2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2$ by $2$, and I got $(2^x-1)x^2 + (2^{x^2-1}-1)x = 2^x -1$. I tried dividing by $2^x-1$ on both sides which would give me a simple quadratic to solve for $x$, but I don't know how to simplify $\frac{(2^{x^2}-1)}{(2^x-1)}$. Am I missing a simple trick here, or am I on a completely wrong path to solving this?
43693
I have to prove that the equation $x^2+y^2+z^2 = n$ only has a solution in the 2-adic integers $\mathbb{Z}_2$ if and only if $n \neq 4^m(8k+7)$, with $k,m \in \mathbb{Z}_{&gt;0}$. In the previous exercises, I already showed that this equation has a solution in $\mathbb{Z}_p$ for all odd primes by using the multivariate Hensel-Rychlik lemma but I have no clue how I could solve this problem.