# If $x+xy+y=55$, what is $x+y$?

If $$x,y\in \mathbb N$$ is s.t. $$x+xy+y=55,$$ find $$x+y$$ ?

I tried to write $$x+xy+y=65$$ as an equation of $$x+y$$, but can't go anywhare :

1. Using the fact that $$xy=\frac{1}{4}((x+y)^2-(x-y)^2,$$ I have that $$x+xy+y=55\iff (x+y)+\frac{1}{4}(x+y)^2-\frac{1}{4}(x-y)^2=65,$$

but I just transposed the problem with $$x-y$$.

Other things I tried is to replace $$xy$$ by $$\frac{1}{2}((x+y)^2-x^2-y^2)$$ but still, I can't do anything with $$x^2+y^2$$.

any idea ?

• @Surb, gotta add $1$ to get that factorization. So, it's divisors of $56$ you want. Commented Jun 17, 2023 at 10:30
• @joshua FYI, please see Simon's Favorite Factoring Trick. Commented Jun 17, 2023 at 10:34
• There are several solutions (three, in fact). For example, $(x,y)=(6,7)$ or $(13,3)$. Can you find the third, using the hints in these comments? Commented Jun 17, 2023 at 10:36
• joshua, what do you mean when you write $\Bbb N$ ? Do you refer to positive integers or nonnegative integers? Commented Jun 17, 2023 at 10:58

Disclaimer: I am going to assume that $$\mathbb{N}$$ includes $$0$$ since @joshua did not specify.

$$x+xy+y=55 \implies x+xy+y+1=(x+1)(y+1)=56$$

Since $$x, y\in\mathbb{N}$$, $$x+1, y+1\in\mathbb{Z}^{+}$$.

Listing out all the cases, we get that \begin{align*} x+y&=0+55=55 \text{ or}\\ &=1+27=28 \text{ or}\\ &=3+13=16 \text{ or}\\ &=6+7=13 \end{align*}

Hence, $$\boxed{x+y=55, 28, 16 \text{ or } 13}$$.

• $x=0$ is wrong as $x\in\mathbb{N}.$ Other than that, good solution. Commented Jun 17, 2023 at 11:02

Since the OP did not state exactly what he means by the symbol $$\,\Bbb N\,$$ (positive integers or nonnegative integers), in my answer I am assuming that $$\,x\,$$ and $$\,y\,$$ are nonnegative integers.

As Gerry pointed out, $$\,x+xy+y=55\,,\,$$ is equivalent to
$$\color{blue}{x+1}+\color{red}{xy+y}=56\quad$$ which is equivalent to
$$\color{blue}{1(x+1)}+\color{red}{y(x+1)}=56\quad$$ which is equivalent to
$$(x+1)(1+y)=56\quad$$ which is equivalent to
$$(x+1)(y+1)=56\,,\quad$$ consequently,

$$x+1=1\quad$$ and $$\quad y+1=56$$

or

$$x+1=2\quad$$ and $$\quad y+1=28$$

or

$$x+1=4\quad$$ and $$\quad y+1=14$$

or

$$x+1=7\quad$$ and $$\quad y+1=8$$

or

$$x+1=8\quad$$ and $$\quad y+1=7$$

or

$$x+1=14\quad$$ and $$\quad y+1=4$$

or

$$x+1=28\quad$$ and $$\quad y+1=2$$

or

$$x+1=56\quad$$ and $$\quad y+1=1$$

Hence,

$$x+y\in\big\{55,28,16,13\big\}\,.$$

In the case that the OP refers to $$\,\Bbb N\,$$ as the set of all positive integers, it results that, $$\,x+y\neq55\,,\,$$ consequently,

$$x+y\in\big\{28,16,13\big\}\,.$$

• Whether $(x,y)=(55,0)$ is a solution depends on your definition of $\Bbb N$. To make this a well-posed question, you should stipluate $x,y\in\Bbb N_{>0}$ or $x,y\in\Bbb N_{\ge 0}$. Commented Jun 17, 2023 at 10:47
• @TonyK, if you read the beginning of my answer, I wrote that I am assuming that $x$ and $y$ are nonnegative integers. Commented Jun 17, 2023 at 10:49
• There's no need for a definition....$\mathbb{N}$ means natural numbers and they start from $1$ Commented Jun 17, 2023 at 11:05
• Does this really matter so much? If OP doesn't want 0 in $\mathbb N$, they can just ignore it. Commented Jun 17, 2023 at 11:24

$$x+y+xy=55$$

$$x+y+xy+1=56$$

$$(x+1)(y+1)=56$$

$$56=1×56=2×28=4×14=7×8=8×7=4×14=28×2=56×1$$

Possible values of $$(x,y)$$ and $$x+y:$$

• $$(0,55), 55$$
• $$(1,27), 28$$
• $$(3,13), 16$$
• $$(6,7), 13$$
• $$(7,6), 13$$
• $$(13,3), 16$$
• $$(27,1), 28$$
• $$(55,0) , 55.$$

We find four unique values for the sum: $$x+y=13\text{ or }16\text{ or }28\text{ or }55.$$