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Let $x$ and $y$ be integers. If $x+y+2xy=83$, find the value of $x+y$.

I tried to multiply both sides by $x+y-2xy$ but I could never manage to simplify it. Is there a better way to solve this question?

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  • $\begingroup$ One equation two unknowns, Why should there be a single solution. If $x=0$ then $y=83$ and $x+y=83$. If $x=1$ then $1+y + 2y=83$ and $y=\frac {82}3$ and $x + y = 28\frac 13$. And so on.... $\endgroup$ – fleablood Apr 28 at 15:12
  • $\begingroup$ The solution is $ x+y=-85$ $\endgroup$ – callculus Apr 28 at 15:21
  • $\begingroup$ @callculus please show your method on how you got the answer $\endgroup$ – Professor of Stupidity Apr 28 at 15:26
  • $\begingroup$ @ProfessorofStupidity I wait for a reply of the OP. $\endgroup$ – callculus Apr 28 at 15:27
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    $\begingroup$ @fleablood Well, $\frac{82}3$ certainly isn't an integer as given. $\endgroup$ – mrtaurho Apr 28 at 15:36
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$$ (2x+1)(2y+1) = 167 $$ which is prime, so we get $$ (0,83),(83,0),(-84, -1),(-1,-84), $$

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Wolog assume $x \le y$ and $y-x = m\ge 0$ then

$2x + m + 2x(x+m) = 83$ and

$2x^2 + (2+2m)x + (m-83) = 0$

$x = \frac {-(2+2m) \pm\sqrt{4m^2+8m + 4-4(m-83)*2}}4=$

$ \frac {-(2+2m) \pm\sqrt{4m^2 + 668}}4=$

$\frac {-1-m\pm \sqrt{m^2 +167}}2\in \mathbb Z$

So $m^2 + 167 = k^2$ for some non-negative integer, $k$, so

$k^2 - m^2 = (k-m)(k+m) = 167$ but $167$ is prime so $k-m =1$ and $k+m=167$ so $m=83$ and $k = 84$

So $x = \frac {-1-83\pm84}2$

So $x = 0, -84$ and $y =83, -1$.

So $x+y = 83$ or $-85$.

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Since our aim is to find $x+y$, Let $x+y=k$, where $k \in \mathbb{Z}$

We have $$k+2x(k-x)=83$$

So

$$2x^2-2kx+83-k=0 $$

The roots are $$x_1,x_2=\frac{k}{2}\pm\frac{1}{2}\sqrt{(k+1)^2-167}$$

So $$(k+1)^2-167=r^2$$ and $167$ being Prime we get:

$$k+1+r=1$$ $$k+1-r=167$$ OR $$k+1+r=-1$$ $$k+1-r=-167$$

Giving $k=83$ and $k=-85$

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  • $\begingroup$ Your solution shows that there's a generalisation: instead of $83$ take any number $m$ such that $2m+1$ is prime. $\endgroup$ – Michael Hoppe Apr 28 at 16:30
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Let's assume we have a solution $(x,y)$. First we make a substitution $y = x + a$ for some integer $a$. Substituting gives $2x^2 +2x(a+1) + (a - 83) = 0$ which we can think of as a quadratic in $x$.

If a general quadratic $ax^2 + bx + c = 0$ has integer roots then we have that it's discriminant $b^2 - 4ac$ is a square because it appears under the squareroot in the quadratic formula.

So in our case, we have that $4(a+1)^2 - 4 \cdot 2 \cdot (a - 83)$ is a square. Simplifying gives us that $4(a^2 + 167)$ is a square. Suppose $a^2 + 167 = k^2$ for some integer $k$, then $(k-a)(k+a) = 167$. Since $167$ is prime we have $k = \pm 84 $ and $a = \pm 83$. Note that $2k$ is the discriminant of our quadratic.

So to find $x$ we plug $a$ into the quadratic formula. $$ x = \frac{-2(a+1) \pm 2k}{2 \cdot 2} = \frac{-2(\pm 83 + 1) \pm 2 \cdot 84}{4} = 0, 83, -84 \textrm{ or } -1. $$ So now we check the possible solutions $(x,y)$. A trick we can use here it to note that the equation $x+y+2xy = 83$ is symmetric in $x$ and $y$ so the possible values for $y$ are also $0, 83, -84$ and $-1$. Going through the options gives the solutions $(0,83), (83, 0), (-84,-1)$ and $(-1,84)$.

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Will Jagy's approach is obviously the correct one. However, there is another way that does not rely on finding the algebraic factorization:

$y·(2x+1) = 83-x$.

$2x+1 \mid 83-x$.

$2x+1 \mid 2·(83-x)+(2x+1) = 167$.

Now check the factors.

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