Find all pairs $(x, y)$ of integers such that: $$1+2^x+2^{2x+1}=y^2$$ Using congruence modulo 3 and 4, I concluded that $x$ is a even integer, $y=1$ $(mod\ 12)$ but I couldn't use them to solve the problem.
32657
What are the solutions in positive integers of the equation:  $${1+2^x+2^{2x+1}=y^2}$$ I tried to factorize the equation but it didn't help much. Clearly $y $ is an odd integer. Substituting $y =2n+1$, we get  $2^x+2^{2x+1}=(2n)\cdot{(2n+2)}$ $\Rightarrow (2^{x-2})\cdot(1+2^{x+1})=(n)\cdot(n+1)$ Which is the product of 2 consecutive integers. Does it help? I don't know.
31441
How can I find all integers $x≠3$ such that $x−3|x^3−3$?  I tried expand $x^3−3$ as a sum but I couldn't find a way after that.
105982
Find all $x \in (-1, +\infty)$ such that $(x^2 + 4x + 3)^x + (2x + 4)^x = (x^2 + 4x + 5)^x$.  What I have done so far was a substitution $y = x + 2$ which results in a nicer form of the equation:   $(y - 1)^{y - 2}(y + 1)^{y - 2} + 2^{y - 2}y^{y - 2} = (y^2 + 1)^{y - 2}$ and $y &gt; 1$, which can then be rewritten as:  $$ \left( \frac{y^2 + 1}{2y} \right)^{y - 2} - \left( \frac{y^2 - 1}{2y} \right)^{y - 2} = 1 $$ Also, my intuition is that $y = 4$ is the unique solution and initially I wanted to prove that this function of $y$ is injective, but it didn't work either.  Do you have any suggestion on how I can continue?
36462
As the title implies.  I'm proving it by contradiction and I've already ruled out $p \equiv 0$ and $2$ (mod $4$).  I'm having trouble proving that it cannot be $\equiv 3$ (mod $4$). Namely, if $p \equiv 3$ (mod 4), then $m^2 + n^2 \equiv 1$ or $3$ (mod $4$). I know that $m^2 + n^2$ cannot be congruent to $3$ (mod $4$); but what about $1$?  Thank you.
92549
Problem: For what pair of positive integers $(a,b)$ is $3^a + 7^b$ a perfect square.  First obviously $(1,0)$ works since $4$ is a perfect square, $(0,0)$ does not work, and $(0,1)$ does not work, so we can exclude cases where $a$ or $b$ are zero for the remainder of this post.  I have a few observations made but not much room for a full solution.   First, since powers of an odd number are odd, and the sum of two odd numbers is even, so the base of the square must be an even number.  Secondly, the last digit of the powers of $3$ are $\{1,3,7,9 \}$ , whereas the last digits of the powers of $7$ are $\{7,9,3,1 \}$.  From here I am not sure how to proceed and any hints much appreciated. I'm not sure if there a finite amount of pairs or not either.
32048
I'm not familiar with diophantine equations. At most my approaches doesn't give results. I need to solve the following equation $$x^3-2y^3=1$$ Where $x,y,z\in\mathbb{Z}$ I know $x=-1,y=-1.x=1,y=0,$ Aware of any other integer solutions. Prove
102624
The question is : prove that $y^2 = x^3+(x+4)^2$ have no solutions in positive integers $x,y$.  I tried to play with the equation and get to $ x^3 = (y+x+4)(y-x-4)$, if $p|x$ then $p^3 | x^3$ and $ p | y+x+4$ or $ p | y-x-4$ , assume that $p$ divides both then we get that $ p | 2x-8$ and since $p|x$ we get that $ p | 8$ meaning $p=2$, so if we assume that $x$ is odd then $ p^3 | y+x+4$ or $p^3| y-x-4$ but not both and so we can write $y+x+4 = b^3 $ and $ y-x-4= a^3$, but that did not help me solving the problem.
48933
Find all pairs of primes $p,q$ such that $pq \mid 2^p +2^q$.  My attempt  : When either one of them is $2$ then easy case checking gives me set of solutions. But what happens when neither of them is $2$?
75685
The only solutions that I know of till now are $(x,y) = (1,1) \space , (19,7)$. We can note that: $$x^2-x+1 = y^3 \implies (2x-1)^2 = 4y^3-3$$ Thus, if odd prime $p \mid y$, then $(2x-1)^2 \equiv -3 \pmod{p}$ and thus, $-3$ is a quadratic residue. This implies that $p \equiv 1 \pmod{6}$ .  How can we further proceed into this problem?  Note: As mentioned in one of the links in the comments, if we instead replace $x$ by $x+1$, we get $x^2+x+1 = y^3$. Thus, the solutions for this is $(x,y) = (0,1) \space, (18,7)$.
35054
Which of the following primes satisfy the congruence   $$a^{24}\equiv6a+2\pmod{13}$$  1) 41  2) 47  3) 67  4) 83  I am interested in Theorem statement, corollary, or Trick or Logic which solves this problem within one minute. Thank you in Advance
64493
I need to describe all numbers of the form $x^2 + 2y^2$.  So far I've reduced the problem to primes, and showed p=2 satisfies it. I've also shown that any primes mod 5 or 7 can't be a written in this form. How do I proceed to show that it holds for all primes mod 1 or 3? I think I'll need to use quadratic reciprocity somehow (only since it is the topic of the homework). I have a hunch the supplementary quadradic recprocity laws will be important.  Any help would be appreciated. Thanks!
47380
If $x$ is positive integer, prove that for all integers $a$, $(a+1)(a+2)\cdots(a+x)$ is congruent to $0\!\!\!\mod x$.  Any hints? What are the useful concepts that may help me solve this problem?
58995
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.
910
I stumbled across this relationship while I was messing around. What's the proof, and how do I understand it intuitively? It doesn't really make sense to me that the sum of odd numbers up to $2x + 1$ should equal $x^2$.
82511
For a prime $p \equiv 1$ or $3$  (mod $8$), show that the equation $x^{2} + 2y^{2} = p$ has a solution.   I'd like to see an elementary proof.
76046
Suppose $k\in\mathbb N$ and suppose $2^k+1=u^2$ and $2^k-1=v^2$ for some $u,v\in\mathbb N$. I have to prove that the first case implies $k=3$ and the second case implies $k=1$. Also, for the second case it is hinted to use the fact (lemma) that if $n$ is odd, then $n^2\equiv1 \pmod 8$, but I suppose it could be used for the first one as well. My progress so far:     First question:      Suppose $u$ is even, then $u=2t$ for some $t\in\mathbb N$, thus $u^2=4t^2=2^k+1$. Notice that two divides the left hand side of the equation, therefore it also divides the right hand side. Since it divides $2^k$ for any $k\ge1$, it must also divide $1$, contradiction. So $u$ is odd and therefore $u^2\equiv1\pmod 8$. Thus there exists an $m\in\mathbb Z$ such that $u^2=8m+1=2^k +1$. Thus $2^k=8m$, therefore $k\ge3$.   I am stuck at this point, however. I suppose setting $k&gt;3$ and finding a contradiction would be a possible method, but I do not know what $k&gt;3$ actually has to offer to solve this problem. No other method I think of gets me anywhere, either. I guess I could try to show that $2^k+1$ cannot be a square for $k&gt;3$. Perhaps I could use the fact that $u^2-2^k=1$ or that $\gcd(u^2,2^k)=1$.     Second question:      The same reasoning can be applied to show that $v$ has to be even. Therefore, according to the lemma proposed above, $2^k-1=v^2\equiv1\pmod 8$. This implies $2^k\equiv2\pmod 8$. For any $k\ge3$ there exists some $t\in \mathbb N$ such that $2^k=8t$, therefore $2^k\equiv0\pmod 8$, which leads to a contradiction. Checking for $k=1,2$ leads to the conclusion that $k=1$ is the only possibility.   I think I have actually figured out this proof while writing down the progress I already have made, but for the sake of completeness and to check it's correctness I will leave it here.
115015
Find all real solutions for $x$ in $$ 2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 . $$  I noticed that $$2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2=2(2^x−1)(x^2-1)+(2^{x^2}−2)x=0.$$  I know that $2^{n+1}-1$ will always be greater than $2^n-1$ for all positive integers. $2^{n-1}-1$ will always be less than $2^n-1$.  How do I solve this problem?