# If $x+y+2xy=83$, find the value of $x+y$.

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?

• 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.... – fleablood Apr 28 at 15:12
• The solution is $x+y=-85$ – callculus Apr 28 at 15:21
• @callculus please show your method on how you got the answer – Professor of Stupidity Apr 28 at 15:26
• @ProfessorofStupidity I wait for a reply of the OP. – callculus Apr 28 at 15:27
• @fleablood Well, $\frac{82}3$ certainly isn't an integer as given. – mrtaurho Apr 28 at 15:36

$$(2x+1)(2y+1) = 167$$ which is prime, so we get $$(0,83),(83,0),(-84, -1),(-1,-84),$$

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 $$-85$$.

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 $$k=-85$$

• Your solution shows that there's a generalisation: instead of $83$ take any number $m$ such that $2m+1$ is prime. – Michael Hoppe Apr 28 at 16:30

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 $$(-1,84)$$.

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.