I'm trying to solve this equation  $$ 2^n + 1 = k^2 $$ but I found only one solution where  $$ n=k=3 $$ Is there are any other solutions and how to prove if there is no more?
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Is there an easy way of showing that $2^n \pm 1$ is never a perfect power, except for $2^3 + 1 = 3^2 $?  I know that   gives the result directly, but I'm hopefully for a more elementary method.    I can show that it is never a square, except for $2^3 + 1$.  Proof: Cases $n=1, 2, 3$ are easily dealt with. Henceforth, let $n\geq4$.  $2^n -1 \equiv 3 \pmod{4}$ hence is never a square.     If $2^n +1 =x^2$, then $2^n = (x-1)(x+1)$ and both of these are powers of 2. Thus we must have $(x-1) = 2, (x+1) = 4$. This gives the solution of $2^3 + 1 = 3^2$.    Let's now do an odd prime.  Say $2^n +1 = x^p$. Then $2^n = x^p - 1 = (x-1)(x^{p-1}+x^{p-2} + \ldots +1)$, so both terms are powers of 2. We have $ x = 2^m+1$ is odd. But the other term is the sum of $p$ odd numbers, hence is odd. Since this is clearly greater than 1, we get a contradiction.  Say $2^n -1 = x^p$. Then $2^n = x^p + 1 = (x+1)(x^{p-1} - x^{p-2} + \ldots -x +1 )$, so both terms are powers of 2. We have $x = 2^m -1$ is odd. But the other term is the sum of $p$ odd numbers, hence is odd. Since $x^p + 1 \neq x+1$ except for $p=1$, this means that the term is greater than 1. Hence we get a contradiction.
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The question asks to verify that each equation is true for every positive integer n.   The question is as follows:  $$1+ 3 + 5 + \cdots + (2n - 1) = n^2$$  I have solved the base step which is where $n = 1$.  However now once I proceed to the inductive step, I get a little lost on where to go next:  Assuming that k is true (k = n), solve for k+1:  (2k - 1) + (2(k+1) - 1) (2k - 1) + (2k+2 - 1) (2k - 1) + (2k + 1)   This is where I am stuck. Do I factor these further to obtain a polynomial of some sort? Or am I missing something?
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I think I have proved that that a cannot be greater than b (ie a &amp; b must be equal) then using this fact I got a quadratic in terms of a and c, trying out some values I got that (3,3,4) is a solution, but how do I prove this is the only solution or if this is not the only solution then what are the other solutions and how do I find them?
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Find all $x$ such that $$x^6=(x+1)^6.$$   So far, I have found the real solution $x= -\frac{1}{2}$, and the complex solution $x = -\sqrt[3]{-1}$.  Are there more, and if so what/how would be the most efficient find all the solutions to this problem? I am struggling to find the rest of the solutions.
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How to prove that the only integral solutions to the equation $$y^{2}=x^{3}-1$$ is $x=1, y=0$.  I rewrote the equation as $y^{2}+1=x^{3}$ and then we can factorize $y^{2}+1$ as $$y^{2}+1 = (y+i) \cdot (y-i)$$ in $\mathbb{Z}[i]$. Next i claim that the factor's $y+i$ and $y-i$ are co-prime. But i am not able to show this. Any help would be useful. Moreover, i would also like to see different proofs of this question.  Extending Consider the equation $$y^{a}=x^{b}-1$$ where $a,b \in \mathbb{Z}$ and $(a,b)=1$ and $a &lt; b$. Then is there any result which states about the nature of the solution to this equation.
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How to find the sum $$\sum^n_{k=1} (k^2+k+1)k!$$  What I tried as follows :   $$\sum^n_{k=1} (k^2+k+1)k!$$  =$$\sum^n_{k=1} (k^2)k!+ \sum^n_{k=1} (k)k! + \sum^n_{k=1} (1)k!$$  Now we can write $\sum^n_{k=1} (k^2) $ as $$\frac{n(n+1)(2n+1)}{6}$$ but what to do with $k!$ please guide thanks.......
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This question comes from trying to see why 24 is the only non-trivial value of $n$ for which $$1^2+2^2+3^2+\cdots+n^2$$ is a perfect square. To this end, let $m,n \in \mathbb N$ be such that $1^2+2^2+3^2+\cdots+n^2 = m^2$, or $$n(n+1)(2n+1) = 6m^2.$$ When $n=24$ the left hand side is $24\times 25 \times 49$ and there are two things that make it work as a solution: $$7^2+1=2\times5^2$$ $$7^2-1=12\times2^2.$$ We can write these algebraically as $$x^2=2y^2-1$$ $$x^2=12z^2+1$$ and solve them simultaneously (with $x,y,z\in \mathbb N$).    These are instances of Pell's equation and each individually has an infinite number of solutions. How do we show there is only one (non-trivial) value of $x$ that is common to the solutions of both equations?
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Find all possible solutions of $$2^x + 3^y = z^2.$$    My approach.   First I substituted $x = 0$, and got the solution, then for $y = 0$. And for $x &gt;  0$ and $y &gt; 0$ , I just know the Pythagorean triple and got the solution as $(4,2,5)$ and $(4,2,-5)$. Please help me to solve it properly.
100189
Find the number of solutions to the equation $$2011^x+2012^x+2013^x-2014^x=0$$  The answer seems to be zero, but I have no idea why. Please avoid considering complex solutions and other scary things.
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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?
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Consider the equation $a^2+ab+b^2=1$. How many integer pairs are solutions to this?  I found 4 pairs: $a=-1, b=0$; $a=0, b=-1$; $a=1,b=0$; $a=0, b=1$. But the solution says the answer is 6. Which other possibilities am I missing? Please post the answer with solution.
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I realize that there is a similar post to this, but that post included a hint which we were not given. Also regarding that hint, I'm just wondering how someone could find it out for themselves. Here is the related post:  I know that $F_n = 2^{2^n} + 1$ and therefore $\,2^{2^n} \equiv -1 \pmod{F_n}$. But I don't intuitively know the next step to take to show that $\ 2^{F_n}\equiv 2\pmod{\!F_n},\,$ i.e. that  $$2^{2^{2^n} + 1} \equiv 2 \pmod{2^{2^n}+1}$$ So in essence, I would like to know how to solve it without knowing the hint. Thank you.
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Solve the equation $2^x + 5^x = 3^x + 4^x$. I can figure out two special solutions $x=0$ and $x=1$, and I try to prove that they are the only two solutions. However, I find it hard to do so because I can't prove the monotony given there are also exponential in the derivative. Any hints to that?
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I'm use to finding the solutions of linear Diophantine equations, but what are you suppose to do when you have quadratic terms? For example consider the following problem:  Find all solutions in positive integers to the following Diophantine equation $x^2 + 2y^2 = z^2$  I'd usually start by finding the gcd and use some other tricks, but I'm not sure how to approach this type of problem
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Find all integer solutions to $x^2+4=y^3$.  Some obvious solutions are $(x,y)=(\pm2,2)$. Are these the only ones?
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I've gotten to the final step and believe my problem lies within my algebra.  Prove the following:   $1 \times 3 + 2 \times 4 + 3 \times 5 + ... + N(N+2) = \frac{N(N+1)(2N+7)}6$  Here is my show that it is true answer:   $\displaystyle\frac{(K+1)[(k+1)+1][(2(k+1)+7]}6$  Here is where I am stuck:  K(K+1)(2K+7) + 6 (k+1) [(k+1)+2]/6  So all that needs to be done is algebra and moving around the above to show that it is equal to the show that it is true answer above. I just cannot figure it out. I am pretty sure I have the right equation.
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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.