Find all positive integers $x,y,z$ such that $$x!+y!=z!$$ My Progress: let $x=y=1$ so that you get $2!$ Which is true. How do I compute this algebraically instead of substituting random values?
124153
I have the formula $x! + y! = z!$ and I'm looking for positive integers that make it true. Upon  inspection it seems that x = y = 1 and z = 2 is the only solution. The problem is how to show it.   From the definition of the factorial function we know $x! = x(x-1)(x-2)...(2)(1)$  So we can do something like this:  $$ [x(x-1)(x-2)...(2)(1)] + [y(y-1)(y-2)...(2)(1)] = [z(z-1)(z-2)...(2)(1)]$$   we can then factor all of the common terms out on the LHS.  $$ [...(2)(1)][x(x-1)(x-2)... + y(y-1)(y-2)...] = [z(z-1)(z-2)...(2)(1)]$$ and divide the common terms out of the right hand side  $$[x(x-1)(x-2)...] + [y(y-1)(y-2)...] = [z(z-1)(z-2)...]$$  I'm stuck on how to proceed and how to make a clearer argument that there is only the one solution (if indeed there is only the one solution).   If anybody can provide a hint as to how to proceed I would appreciate it.
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!
5211
Show that if $x$, $y$, $z$ are integers such that $x^2 + y^2 = z^2$, then at least one of them is divisible by $2$, at least one is divisible by $3$, and at least one is divisible by $5$.  I know that $x,y,z$ are sides of a right-angled triangles. If I start putting values, then I can show, but have no clue about mathematical approach.  How do I solve it mathematically, using modular arithmetic?
67835
The equation is $(x!)(y!) = x!+y!+z! $  where $x,y,z$ are natural numbers.  How to find out them all?
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.
72479
Find all triples of positive integers (a, b, c) such that  $$ \left( 1 + \frac{1}{a} \right)\left( 1 + \frac{1}{b} \right)\left( 1 + \frac{1}{c} \right) = 2 $$  I tried to expand, yielding $ a+b+c+ab+bc+ca+1 = abc $, but I couldn't go anywhere from here.   This problem clearly involves using the fact that $a, b, c$ are positive integers, since there are three variables and only one equation. It would be great if someone also addressed this condition and maybe tried to solve the problem for other requirements, like when $a, b, c$ are positive or negative integers.  Finally - can anyone post or link other such problems which involves a mix of algebra manipulation and number theory. I would really appreciate it.  (Here is the full paper: )
84382
I know the answer is obvious: In $\mathbb{Z}$ the only solutions of $xy=-1$ are $x=-y=1$ and $x=-y=-1$. My problem is that I want to formally prove it and I don't know how to write it. Where do you even begin for such a trivial statement?  Edit: I would like to prove it viewing $\mathbb{Z}$ as a ring. This is, just using the sum and product of integers. No order, no absolute value, etc...
45370
My teacher in Math Team gave the following question to us.     Solve $$a!b!=a!+b!+c!$$ where $a$, $b$ and $c$ are nonnegative integers.   I found only one solution by trial and error and it is $(a,b,c)=(3,3,4)$.  Any help is appreciated!
75196
I am trying to find a way for the positive integers written as the sum of other positive integers.( expressed in terms of some functions)  I searched a bit and I came across with  But in my case the order matters. What I mean is for example if we write for 4,  there would be 7 ways to express 4 as;           1 + 3           2 + 2           3 + 1           1 + 1 + 2           1 + 2 + 1            2 + 1 + 1           1 + 1 + 1 + 1   Can anyone guide me in approaching to the solution of my problem? Thanks in advance.
115559
I have been trying to show these two inequalities hold for all positive integers n, but I don't know how to proceed at all... I have tried playing around with them but I didn't find anything helpful. I have also tried induction but I couldn't make it work.
98209
The equation $X^{n}  + Y^{n}  = Z^{n}$    , where $n   \geq  3$     is a natural number, has no solutions at all where $X,Y,Z$     are integers.   My solution:  False. Because if we let $X=0  ,Y=0$,  then $0^{3}+0^{3}=0^{3}$  . Thus there is a solution for the equation $X^{n}  + Y^{n}  = Z^{n}$,   where $X,Y,Z$   are integers.  Can anyone please give feedback on my answer and state whether it is correct on not.
29391
find all  positive integers $a,b,c$ such that  $abc=24$  $ab+bc+ca=38$  If particular values are given then we can easily find the solution but I am searching for some short general method.  Is there any sufficient condition for the constant values to have integer solution?
106987
Find all positive integers $a,b$ such that each of the equations $$x^2-ax+b=0$$ and $$x^2-bx+a=0$$ has distinct positive integral roots.   Suppose $\alpha,\beta$ are roots of the first equation and $\gamma,\delta$ are of the second, then we have these two sets of equations  $$a=\alpha+\beta$$ $$b=\alpha\beta$$ and $$b=\gamma+\delta$$ $$a=\gamma\delta$$ How to proceed?
81791
Determine the set of points $z$ that satisfy the condition $|2z|&gt;|1+z^2|$  I tried to redo this problem and got to this point  $|2z|&gt;|1+z^2|$ $\Rightarrow$ $2|z|&gt;1+|z^2|$ $\Rightarrow$ $2|z|&gt;1+z\overline z$  Let $z=x+iy$ then   $$2\sqrt{x^2+y^2}&gt;1+(x+iy)(x-iy)$$  $$2\sqrt{x^2+y^2}&gt;1+(x^2+y^2)$$  $$0&gt;1-2\sqrt{x^2+y^2}+(x^2+y^2)$$  $$0&gt;(1-\sqrt{x^2+y^2})^2$$  since $x,y$ are real number , so there is no solution for this inequality?
24510
Solve for integers $x, y, z$ such that $x + y = 1 - z$ and $x^3 + y^3 = 1 - z^2$.  I think we'll have to use number theory to do it. Simply solving the equations won't do. If we divide the second equation by the first, we get: $$x^2 - xy + y^2 = 1 + z.$$ Also, since they are integers $z^2 \ge z \implies -z^2 \le -z$. This implies $$x + y = 1 - z \ge 1 - z^2 = x^3 + y^3.$$ This shows that atleast one of $x$ and $y$ is negative with the additive inverse of the negative being larger than that of the positive. I have tried but am not able to proceed further. Can you help me with this? Thanks.
22327
Let $u$ be an integer of the form $4n+3$, where $n$ is a positive integer. Can we find integers $a$ and $b$ such that $u = a^2 + b^2$? If not, how to establish this for a fact?
32643
Find all positive integers $a, b, c$ such that $1/a + 1/b + 1/c = 4/3999$.  The contest is just ended, so you may freely answer. (I did not attend the contest: it is an Italian contest for schools and the groups may ask other people for hints. This is how I got involved, but of course asking on MSE would have been cheating)  I found three solutions: I am explaining below what I did. I would like however to know if there is a more general approach.