Use the method of ascent to prove there are infinitely many solutions to the Diophantine equation: $$x^2 − 3y^2 = 1$$ We can do this by showing how, given one solution $(u, v)$, we can compute another solution $(w, z)$ that is larger is some suitable sense. Then my proof will involve finding a pair of formulas, something like: $w = x + y$ and $z = x − y$. However I tried these formulas and they don't work. So I asked my teacher and she said that there is a pair of second degree formulas which do work; one of them has a cross term and one of them involves the number 3.
78853
I have seen the Pell's equation wiki page but I need to prove this from scratch without mentioning any formula. I have also seen multiple answers on this site but the answers tend to skip over and assume formulas.  This is what people do -> $x^2 - 3y^2 = 1$ has solution 1, 0.  Next they say we "observe" that $$(x',y') = (2x + 3y, x + 2y)$$ is also a solution.  What I'm asking is how they get this value. Can someone help?
91139
How many solutions are there to the equation      $x_1 + x_2 + x_3 + x_4 + x_5 = 21$    where $x_1, i = 1,2,3,4,5$ is a nonnegative integer such that $0 ≤ x_1 ≤ 3, 1 ≤ x_2 &lt; 4$, and $x_3 ≥ 15$?  I have correctly completed the previous parts of the question but am having trouble with this particular restriction.  The total number of solutions to the equation is $C(25,21) = 12650$.  I have tried to do each restriction individually:  For $0 ≤ x_1 ≤ 3$, I solved this by subtracting the number of solutions when $x_1 ≥ 4$ from the total number of solutions. This is $C(25,21) - C(21,17) = 5985$.  For $1 ≤ x_2 &lt; 4$, I have $C(24,20) - C(21,17) = 5985$.  For $x_3 ≥ 15$, I have $C(10,6) = 210$.  However, the restriction is when these are all together. I am not sure how to proceed. The answer is $106$.
76358
How to solve the Diophantine equation $x^2 + 2 = y^3$ with $x,y&gt;0$ ? ($x,y$ are integers.)
84034
I need help solving this past exam question, the professor posted exam questions with out solutions to last year's exam, I tried reduction of order, however that seemed to complicate things. And then I tried Euler-Cauchy method however I realized that won't work due to the $(2x^2 + 1)y''$.  Given $y_1=x$ is a solution $$(2x^2 + 1)y'' − 4xy' + 4y = 0$$ Find the second solution $y_2$ for equation.
71283
$$ 3x+6y+5z=7 $$ The general solution to this linear Diophantine equation is as described    $$ x = 5k+2l+14 $$ $$ y = -l $$ $$ z = -7-k $$ $$ k,l \in \mathbb{Z} $$  If I plug the original equation into  the solution is: $$ y = 5n+2x+2  $$ $$ z =-6n-3x-1 $$ $$ n \in \mathbb{Z} $$  I can rewrite this as:  $$ x = l $$ $$ y = 5k+2l+2 $$ $$ z = -6k-3l-1 $$ $$ k,l \in \mathbb{Z} $$  However now two equations depend on two variables ($k,l$) and one on one variable $l$. In the first solution one equation depends on two variables and two on one variable.  Questions:  How can I come from a representation like the one from wolfram alpha for the general solution to one where all equations depend on one distinct variable except one equation.  Is there always such a representation?
25330
The full question is to solve $f(x)f(2x^2) = f(2x^3+x)$, prove there are at most one solution per degree of $f(x)$ and then find all the solutions. (Here $f(x)$ is assumed to be polynomial.)  So far I have proven that $(x^2+1)^n$ works for even degree polynomials, and I'm pretty near certain there are no odd solutions.  Is the following a valid argument for there only being at most 1 solution per degree? Let $K$ be the degree of $f(x)$ then $f(x)f(2x^2) = f(2x^3+x)$ produces $3k+1$ linearly independent equations with only $k+1$ independent variables. Therefore for any degree $K$ polynomial there can be at most 1 solution.  Finally, is there any elegant way to prove that no odd solutions exist?
62629
Find one integer solution to the following Diophantine equation:   \begin{equation*} \\\forall \,x,y \in \mathbb{Z}\\ 6xy + x - y = 274 \end{equation*}   The resultat ist $x = 9$, $y = 5$ (Obtained with  )  I need to know a method that allows me to reach such a result. How can I calculate the value of these two variables in a Diophantine equation? For example: 6xy + x - y = 458
65459
One way to prove that there is an infinity of consecutive powerful integers is to just note that there are infinite solutions to the equation $x^2-8y^2=1$, because clearly $x^2$ and $8y^2$ are clearly powerful. But this uses a bit of theory.  I found this problem under an induction worksheet, and all of the surrounding problems are easy. Is there really an elementary solution to the problem?  All of my attempts have been fruitless.  Note: a number is powerful if whenever a prime $p$ divides it we have that $p^2$ also divides it.
106183
Find the number of solutions of $x^2=x\mod m$ for any integer $m$.  I was thinking about bringing the $x$ over to the other side and somehow apply chinese remainder thm.
30555
What are the integer solutions to the diophantine equation   $$n^2 + 1 = 2 \times 5 ^m? $$  We have $(n,m) = (3,1), (7, 2) $ as solutions. Are there any more?    This seems like it would be a well known diophantine equation, but I can't seem to find any information about it.
117536
Prove that the equation $$x^2 - x + 1 = p(x+y)$$ has integral solutions for infinitely many primes $p$.  First, we prove that there is a solution for at least one prime, $p$. Now, $x(x-1) + 1$ is always odd so there is no solution for $p=2$. We prove there is a solution for $p=3$. If $p=3$, $y = (x-2)^2/3-1$. We get integral solutions whenever we get $x = 3m +2$, where $m$ is any integer.$\\$ We provide a proof by contradiction which is   similar to Euclid's proof of there being infinitely many primes. \Let us assume that it is is true for only finitely many primes, and name the largest prime for which the equation is true $P$.\\We set $$x = 2\cdot3\cdot5\dots P$$ $x$ is the product of all primes upto $P.\\$ Then, the term $x(x-1) + 1$ is either prime or composite. If it is prime, then we set $p = x(x-1) + 1, y = 1 - x$ and get a solution. If it is composite, we write $x(x-1) + 1 = p\times q$, where $p$ is any prime factor of $x(x-1)+1$, and $q$ is an integer, $q = (x(x-1)+1)/p$. Now, $x(x-1) + 1$ is not divisible by any prime upto $P$ since it leaves a remainder of $1$ with all of them. So, $p &gt; P$. We set $y=q-x$ for a solution.$\\$In either case, we get a solution for a prime $p &gt; P$, which means there's no largest prime for which this equation has solutions. This contradicts the assumption that there are solutions for only finitely many primes.  I feel like I'm missing some step. Is this correct ?
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.
86256
How can I count all integer solutions to the equation $$x^2+dy^2=n$$  given $n$ and $d$, where $n$ is composite (or prime)?  By counting I mean any algorithm faster than brute force enumeration.
26918
I am assigned to find solutions to the Pell-type equation $x^2-2y^2=-1$. So far, I only have $(7,5), (41,29)$ and $(239,169)$.  My question is, is there a general formula to find all its solutions? Are there infinitely many solutions? If it's possible to write Mathematica code that would find its solutions, what would the code look like?  Thanks a lot.
105162
Suppose $n = pq$ with $p$ and $q$ distinct odd primes.   Suppose that $\gcd(a,pq)=1$. Prove that if the equation $x^{2} \equiv a\pmod n$ has any solutions, then it has four solutions.   Proof: Suppose $n = pq$ with $p$ and $q$ distinct odd primes and that gcd($a,pq$) = 1. Let us have the equation $x^{2} \equiv a\pmod n$. Then, $x^{2} \equiv a$ (mod $pq$), $p \not = q$. By definition, we can separate the equation into two equations such that $y^{2} \equiv a \equiv b$ (mod $p$) and $z^{2} \equiv a \equiv c$ (mod $q$). Let $g_{p}$ be a primitive root modulo $p$ and $g_{q}$ be a primitive root modulo $q$. Then, $b$ is equal to some power of $g_{p}$ and $c$ is equal to some power of $g_{q}$. With the fact that $b$ has a square root modulo $p$ (i.e. $r^{2} \equiv b$(mod $\ p)$) and $c$ has a square root modulo $q$ (i.e. $t^{2} \equiv c$(mod $\ q)$), there is an even power of $g_{p}$ and of $g_{q}$ such that $b = g_{p}^{2k_{1}}$(mod $\ p)$ and $c = g_{q}^{2k_{2}}$(mod $\ q)$ for some $k_{1}, k_{2} \in Z$. By computing, we have the following:  $r^{2} \equiv b$(mod $\ p)$       $\equiv b^{(p + 1) / 2}($mod $\ p)$       $\equiv (g_{p}^{2k_{1}})^{(p + 1) / 2}($mod $\ p)$       $\equiv (g_{p}^{p + 1})^{k_{1}}($mod $\ p)$       $\equiv g_{p}^{2k_{1} + (p - 1)k_{1}}($mod $\ p)$       $\equiv b \cdot g_{p}^{(p - 1)k_{1}}($mod $\ p)$       $\equiv b$ and $t^{2} \equiv c$(mod $\ q)$        $\equiv c^{(q + 1) / 2}($mod $\ q)$        $\equiv (g_{q}^{2k_{2}})^{(q + 1) / 2}($mod $\ q)$        $\equiv (g_{q}^{q + 1})^{k_{2}}($mod $\ q)$        $\equiv g_{q}^{2k_{2} + (q - 1)k_{2}}($mod $\ q)$        $\equiv c \cdot g_{q}^{(q - 1)k_{2}}($mod $\ q)$        $\equiv c$  Hence, $r$ is a square root of a modulo $p$ and $t$ is a square root of a modulo $q$, which means there are two solutions for each $p$ and $q$. Since $p \not = q \in \mathbb{Z}_{n}$, we have isomorphism such that $\mathbb{Z}_{n} \simeq \mathbb{Z}_{p} \cdot \mathbb{Z}_{q}$. Therefore, if the equation $x^{2} \equiv a$ (mod $n$) has any solutions, it must have four solutions. $\blacklozenge$  What do you think about the proof I wrote?
122307
Given $a,b,c,d \in \mathbb{Z}$, there is $x,y$ such that  $$(a^2 + b^2)(c^2 + d^2) = x^2 + y^2$$  One can show this by considering the complex number $a + bi$ and $c+ di$, using complex properties to deduce that $x = ac - bd, y = ad + bc$ is a solution.   However, given that either $a^2, b^2$ are distinct or $c^2,d^2$ are distinct, then how can one can find nonzero solutions $x,y$?  Sorry, I am assuming $a,b,c,d$ are nonzero.
28243
If $x,y \in \mathbb Z$ , then find all the solutions of   $$y^2=3x^4+3x^2+1$$       I was asked this question by my friend who said that he encountered this while solving another problem. I have tried several things but am unable to solve this question. Moreover, this has to be done using elementary methods only. So far, I have tried to factorize and use Pell's equation. At the end, I'm getting $$2y_{n} + (2x^2_{n}+1)\sqrt{3}=(2+\sqrt{3})^{n}$$   where $n \in \mathbb Z^{+}$    But I'm not able to figure out how to show a contradiction from here. Can anyone please help me out? Thanks.