Consider the following Diophantine equation $$x^2 + xy + y^2 = n\,.$$ For a particular positive integer $n$, the number of solutions $\left(x,y\right)$ such that $x$ and $y$ are integers is given by the function $S(n)$. The function $S(n)$ is not one-to-one. For example, each number $n$ in the set $$\{1, 4, 9, 16, 25, 36\}$$ corresponds with $S(n) = 6$. In increasing order, starting from $n = 1$, the first $n$ such that $S(n) = 36$ is $637$.  What is the $500$th $n$ such that $S(n) = 36$?  Any precise approach or hint from where I should start the problem?
121137
Let $S$ be the set of integers $n$, such there exist integers $x,y$ with $$n=x^2+xy+y^2$$   Is the implication $$a,b\in S\implies ab\in S$$ true? If yes, how can I prove it?  I worked out $$n\in S\iff 4n\in S$$ and $$n\in S\iff 3n\in S$$ I tried two approaches. The first is to express $$(a^2+ab+b^2)(c^2+cd+d^2)$$ in the form $$f^2+fg+g^2$$ with polynomials $f,g$ with integer coefficients. I however could not find suitable $f$ and $g$. The second approach is based on $$x^2+xy+y^2=\frac{(2x+y)^2+3y^2}{4}$$ If we have $n=x^2+xy+y^2$ , we have $u^2+3v^2=4n$ for some integers $u,v$ with equal parity. The main problem of this approach is to consider the equal parity. Any ideas ?
14071
I want to find a set of integer solutions of Diophantine equation: $ax + by = c$, and apparently $\gcd(a,b)|c$. Then by what formula can I use to find $x$ and $y$ ?   I tried to play around with it: $x = (c - by)/a$, hence $a|(c - by)$.     $a$, $c$ and $b$ are known. So to obtain integer solution for $a$, then $c - by = ak$, and I lost from here, because $y = (c - ak)/b$. I kept repeating this routine and could not find a way to get rid of it? Any hint?  Thanks, Chan
120715
Let $a\in \mathbb{Z}/p\mathbb{Z}$ and $m$ is an odd number. Then show that the number of solutions to $x^2\equiv a\pmod m$ is $$\prod\limits_{p|m}\left(1+\left({\tfrac {a}{p}}\right)\right)$$  N.B I know I have to use CRT but how to get the expression?
103470
Find all integers $$(x, y)$$ such that $$1 + 2^x + 2^{2x + 1} = y^2$$ So I basically used $$ f(x) = 1 + 2^x + 2^{2x + 1} = y^2$$ and created a table from 0 to 20. I got two pairs of integers: $$(0, \pm2)$$ and $$(4, \pm23)$$ I want to know if there's another method or if there are any other pairs of integers, or a generalisation.
79310
The problem is what are the integer solutions to  $$7x^2 - 40xy + 7y^2 = (|(x - y)| + 2)^3$$  I only got $x - y = 4 (\mod 13)$. Also, how to solve it?
15580
Prove that $x^2 + y^2 = z^4$ has infinitely many solutions with $(x,y,z)=1$.   Do I use the terms $x= r^2 - s^2$, $y = 2rs$, and $z = r^2 + s^2$ to prove this problem?  Thanks for any help.
112528
If I have the Diophantine equation $\displaystyle{\sum_{i=1}^n x_i =A}$, is there a function $f(n,A)$ that will yield the number of non-negative integer solutions of the equation?
13712
I have been reading about this quadratic Diophantine equation of the form  $x^2 + axy + y^2 = z^2$  where x, y, z are integers to be solved and a is a given integer.  All integral solutions are given by  $x = k(an^2 - 2mn), y = k(m^2 - n^2), z = k(amn - m^2 - n^2)$ and  $x = k(m^2 - n^2), y = k(an^2 - 2mn), z = k(amn - m^2 - n^2)$  (due to diagonal symmetry in x and y)  where $m,n$ are integers with $\gcd(m,n) = 1,$ but $k \in \mathbb Q$ is rational such that $(a^2 - 4) \, k \in \mathbb Z.$ This is Theorem 2.3.2. on page 90 of An Introduction to Diophantine Equations by Andreescu, Andrica, and Cucurezeranu. (2010). EDIT BY WILL JAGY.  I have no problem understanding how the solution forms were derived; they were just basic algebraic manipulation. But then when it comes to the solutions in positive integers, the form becomes  $x = k(an^2 + 2mn), y = k(m^2 - n^2), z = k|amn + m^2 + n^2|$ and  $x = k(m^2 - n^2), y = k(an^2 + 2mn), z = k|amn + m^2 + n^2|$  where k, m, n are positive integers, an + 2 m > 0 and m > n.  What I can understand is that we apply modulus to the x, y, and z in the previous form to get the latter form (we want x, y, and z to be in positive integers), but I can't seem to understand how an + 2 m > 0 and m > n work to prove  $|x| = |k(an^2 - 2mn)| = kn|an - 2m| = kn(an + 2m) = k(an^2 + 2mn)$ and  $|z| = |k(amn - m^2 - n^2)| = k|amn + m^2 + n^2|$.  Can anyone help me on this? I've been pondering for almost a week. It's driving me crazy. Thank you in advance.
76358
How to solve the Diophantine equation $x^2 + 2 = y^3$ with $x,y&gt;0$ ? ($x,y$ are integers.)
113082
I've heard a famous result that $26$ is the only integer, such that $26-1=25$ is a square number and $26+1=27$ is a cubic number.In other words, $(x,y)=(5,3)$ is the only solution for $x^2+2=y^3$.  How if we make it like this $x^3+2=y^2$? Are there any integral solutions? If so, finite or infinite many?  I've checked first $100$ naturals and no solutions satisfy the equation. However, I have no ideas how to start the proof.
23395
So, a question in Elementary Number Theory by David M. Burton:     Problems 4.4      2(a) Using congruences, solve the diophantine equation 4x + 51y = 9       Hint: $4x \equiv 9(mod \ 51) $ gives $ x = 15 +51t$, whereas $ \ 51y \equiv 9 (mod \ 4)$ gives $y = 3+4s $. Find the relation between $s$ and $t$ From this hint, I substitute $ x = 15 +51t$ and $y = 3+4s$ in the given equation $4x + 51y = 9$ to get the relation $s = -t-1$ and using which I can get the solutions $ x = 15 + 51t, \ y =-1-4t$ But we know that the solutions of $ax \equiv b (mod \ n)$ is equivatent to the solutions of the diophantine equation $ax -ny = b$, so, solutions of $4x \equiv 9(mod \ 51) $ is equivalent to solutions of the equation $4x-51y=9$ whereas, solutions of $51y \equiv 9(mod \ 4) $ is equivalent to solutions of the equation $-4x+51y=9$ Now, my question is: as, the above equations $4x-51y=9$ and $-4x+51y=9$ are different, then what is the point of substituting $ x = 15 +51t$ (solution of $4x-51y=9$) and $y = 3+4s $ (solution of $-4x+51y=9$) in the given equation $4x+51y=9$ to find a relation between $t$ and $s$ ? Isn't it absurd as $ x = 15 +51t$ and $y = 3+4s $ are solutions of two different equations? Can anyone please explain this? PS: I did it like this $4x \equiv 9(mod \ 51) $ gives $ x = 15 +51t$ and by substituting it in $4x+51y=9$ we get $y=-1-4t$
75008
Consider the polynomial $x^2+x=0$ over $\mathbb Z/n\mathbb Z$   a)Find an n such that the equation has at least 4 solutions  b)Find an n such that the equation has at least 8 solutions  My idea is to check individual n, and I found the answer for a is n=6 at x={0,2,3,5} f(x)={0, 6, 12, 30)  Is there any shorter way?
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 ?
75654
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
87594
How to list all integer solutions of   $m^2-n^5 = m - n$  Here $m$ and $n$ are some positive integers.  Also, I want to know the name of this type equations (if name exist).   Regards Rosy
56779
Find the number of integer solutions to $x_1 + x_2 + x_3 + x_4 = 25$ with $ 1 \leq x_1 \leq  6, 2 \leq x_2 \leq 8, 0 \leq x_3 \leq 8, 5 \leq x_4 \leq 9.$  Firstly, I defined $y_i = x_i - lower bound$ so for example $y_1 = x_1 - 1$ This gives $y_1 + y_2 + y_3 + y_4 = 17$ with $$0 \leq y_1 \leq 5, 0 \leq y_2 \leq 6, 0 \leq y_3 \leq 8, 0 \leq y_4 \leq 4$$ Next I applied Inclusion-Exclusion Principle;  Total number of ways = $17+3 \choose 3$  Number of ways with just $y_i$ breaking upper bound;  $y_1;$$11+3 \choose 3$  $y_2;$ $10+3 \choose 3$  $y_3;$ $8+3 \choose 3$  $y_4;$ $12+3 \choose 3$  Similarly I calculated the number of ways with $y_i$ and $y_j$ breaking upper bound at the same time. And since there is no way in which 3 of them break their upper bounds simultaneously I was left with ;  $${20 \choose 3} - {14 \choose 3} - {13 \choose 3} - {11 \choose 3} - {15 \choose 3} + {7\choose 3} + {5 \choose 3} + { 9 \choose 3} + {4 \choose 3} + {8 \choose 3} + {6 \choose 3} $$  as my final answer.  Is this solution correct and is there a nicer way of solving it/ any insights about this type of problem?  Note, this problem is from 'An Introduction to Combinatorics and Graph Theory'
81147
I need help interpreting this passage from page 35 of the book Diophantine Analysis by Robert D Carmichael.  "Thus the problem of solving Eq. (1) is reduced to that of solving Eq. (4) for $u$ and $v$ and choosing one of those values for which x and y have the desired characteristic.  But the problem of solving Eq. (4) is identical with that of the representation of a given integer by means of a binary quadratic form.  The plan of this book does not permit the detailed development of this latter subject."  Equation 1 is $ax^2+2bxy+cy^2+2dx+2ey+f=0$ and equation 4 is $au^2+2buv+cv^2=m$.  $u,v,$ and $m$ are defined elsewhere if that matters, but what I need to know is what that bold part is referring to; and that last sentence gives me little reason to believe I can find context in the book to make sense of it.  This is the main part of the book I'm interested in, but I do intend to go through most of it.  It wasn't easy to find a book on the subject and I just checked it out from the library today.