Prime number equation

The number of solutions of the equation $xy(x+y)=2010$ where $x$ and $y$ denote positive prime numbers, is ____

I tried various things but nothing seems to work out. $2010$ can be resolved into $67\times30$ but $30$ is not a prime number. Moreover, they do not come under the format $xy(x+y)$. Please help me.

• Hint: since $67$ is a prime number, then $67\mid x$, $67\mid y$ or $67\mid (x+y)$ – Michael Galuza Jul 20 '15 at 9:49
• hm how will this helps? – Dr. Sonnhard Graubner Jul 20 '15 at 9:57
• Your title is misleading. – Dietrich Burde Jul 20 '15 at 9:58

HINT:

WLOG $x\le y$

$\implies 2010\ge x^2(x+x)\iff x^3\le1050$

and $11^3=1331\implies x<11$

Alternatively,

If $67|x, x=67m$ where integer $m>0$

$\implies2010=67my(67m+y)\iff 30= my(67m+y)$

But $67m+y>67$

$\implies67|(x+y)\implies x+y=67n$ where integer $n\ge1$

$\implies xy=\dfrac{30}n$

But $(x+y)^2-2xy=(x-y)^2\ge0$

$\implies(67n)^2\ge2\cdot\dfrac{30}n\iff n^3\ge\dfrac{2\cdot30}{67^2}<1$

• $x$ does not have to be prime – Rol Jul 20 '15 at 10:08
• I think you have the wrong equation. $xy(x+y)=2010$. $xy(x+y)+1 \neq (x+1)(y+1)$. – wythagoras Jul 20 '15 at 11:02
• @wythagoras. thanks – lab bhattacharjee Jul 20 '15 at 11:09

First, let's factorize $2010$: $$2010=2\cdot 3\cdot 5\cdot 67$$ If $67\in\{x,y\}$, say WLOG that $x=67$, so $$\frac{2010}{xy}=x+y$$ or $$\frac{30}y=67+y$$ which is impossible.

If not, $67$ is a factor of $x+y$ and $67<x+y<xy<15$.

Thus, there is no solution.

Every prime number more than 3 is in the form 6k + 1 or 6k - 1

There are these conditions :

x = 1 (mod 6) , y = 1 (mod 6) : xy (x + y ) = 2 (mod 6) but 2010 = 0 (mod 6)

x = -1 (mod 6) , y = 1 (mod 6) : xy = 0 (mod 6) so xy=30 , (x + y ) = 67

x = -1 (mod 6) , y = -1 (mod 6) : xy (x + y ) = 4 (mod 6) but 2010 = 0 (mod 6)

x = 2 y = p : 2p( 2 + p ) = 2010 with no answer

x = 3 y = p : 3p( 3 + p ) = 2010 with no answer

Therefor there is no answer to this equation.