Find all integer solutions of the equation $a^2+ b^2 + c^2=a^2 b^2$.  This is one of the questions we presented in one session to contest preparation PUTNAM. It turns out that I can't get from the problem. Could someone just give me a hint? (Please, don't give me the answer. Simply, an argument that can help me advance in the problem or theorem might suffice.)
113894
I am unfamiliar with this type of problem. How does one solve this and under what category of math does this fall under.  Find the integral solutions for $x^2+y^2+z^2=x^2y^2$
12531
$m^2+(m+1)^2=n^4+(n+1)^4$  Integral (integer) solutions needed I could find only four, are there others  -1 &amp; 0 give both sides as 1.
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?
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Generally one can see that there are infinite number of solutions for this equation $$a^{2}+b^{2}=c^{2}$$ by taking multiples of the solution $3,4$ and $5$.   Can i use this as a fact to prove, that $k^{2} + (k+1)^{2}$ is a perfect square for infinitely many $k \in \mathbb{N}$? Any hints, suggestions would be helpful. If not, then how do i prove this fact!
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I want to solve ${y^2}=x^4+x^3+x^2+x+1$ in $\mathbb{Z}$.    I can find four solutions. Is there another solution? I know that there are six solutions. $$(-1, \pm 1)\;,\;(0, \pm 1).$$  My try: \begin{align}   &amp; x^4+x^3+x^2+x+1=0\quad \Rightarrow \underbrace{\left( x+\frac{1}{x} \right)^2}_{t^2}+\underbrace{\left( x+\frac{1}{x} \right)}_{t}-1=0\,\Rightarrow t\notin \mathbb{Z} \\   &amp; x^4+x^3+x^2+x+1={\pm 1}^2\quad\Rightarrow \,x=-1\,\,,\,x=0.\\  \end{align}   Thanks.
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The problem from Burton: show that the equation $n^2 + (n+1)^2 = m^3$ has no solution in the positive integers.  So far, I can see that gcd($n$,$n+1$)$=1$ and $m \equiv_4 1$ and $m=a^2 + b^2$ for some integers a,b. I'm guessing I need to reach a contradiction.  At this point, I am stuck. Any hints?
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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.
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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?
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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.
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Find one integer solution to the Diophantine equation   \begin{equation*} 18x+14y+63z=5. \end{equation*}   If this were only a linear equation over $\mathbb{Z}^2$, then I could easily solve it by using the extended Euclidean algorithm... but I have no idea how to do this with more than 2 unknowns...
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This question concerns the equation $$\frac{a}{a^2+1} + \frac{b}{b^2+1} = \frac{c}{c^2+1}$$ and the possibility of rational number solutions with $abc \ne 0$.  In comments arising from:  It is mentioned that this equation has an infinite number of such solutions.  How can we determine that there are an infinite number of such solutions?  And is there a way to generate these solutions (or at least find a few of them)?
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The solutions to the equation $z^2-2z+2=0$ are $(a+i)$ and $(b-i)$ where $a$ and $b$ are integers.  What is $a+b$?  I simplified and got $(z+1)(z+1) = -1$ and now I'm not sure where to go from there. I did this but I'm not sure.  $(a+i)^2 = a^2 - 1$  $(b-i)^2 = b^2 + 1$  $a+b=(a+i)+(b-i)=(a^2-1)+(b^2+1)=a^2+b^2$
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Find all integer roots of: $x^2(y-1)+y^2(x-1)=1$   Obviously $(2,1)$ and $(1,2)$ are two answers. But I was unable to manipulate the equation algebraically giving a useful form for finding all other possible answers! I also tried to view it as a quadratic equation in terms of $x$, but forming the Delta of that equation didn't help much other than if $y=1$ then certainly Delta would be a square and equation would have answer for $x$...
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How to solve the Diophantine equation $x^2 + 2 = y^3$ with $x,y&gt;0$ ? ($x,y$ are integers.)
123904
We had received some questions on Quadratic equations, But I wasnt able to do one. Here it goes:  Let $a,b$ be natural numbers $a&gt;1$. Also, $p$ is a prime number. If $ax^2+bx+c=p$ for 2 distinct integral values of $x$. Then the number of integral roots of the equation $ax^2+bx+c=2p$ is ? Well I know the answer is $0$, But I am not able to get it properly.
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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
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I know how to solve $ax + by = c$ using Extended Algorithm. But with more than variables, I'm lost :(.   To verify if it has an integer solution is easy, since we only need to check for $\gcd(a,b,c)|d$. Other than that, how can we find an integer solution for this equation?    Thanks, Chan