Seperable Integers

Let $n=a(a+2)=b(b+9)$ be a number that is both two-separable and nine-separable. Since $a(a+2)=b(b+9)$ , we can rewrite this as

$77=(2(b-a)+7)(2(b+a)+11)$

Since the second term is nonnegative, the first term must be nonnegative. Furthermore, $2(b-a)+7\le2(b+a)+11\iff a\ge-1$ , which is true, so $2(b-a)+7\le2(b+a)+11$ . Therefore, $2(b+a)+11=11$ or $2(b+a)+11=77$ . Since the first case implies $b+a=0$ , the second case is the only possible case. It suffices to solve the system of equations

$\begin{cases} 2(b+a)+11=77\\ 2(b-a)+7=1 \end{cases}$

whose solution is $(a,b)=(18,15)$ . Therefore, the only integer that is both two-separable and nine-separable is $n=18(18+2)=15(15+9)=\boxed{360}$ .

Suppose $n$ is an integer which is both two-separable and nine-separable then $n=a(a+2)=b(b+9)$ where $a$ and $b$ are positive integers.

If $b\ge a$ then $b(b+2)\ge b(b+9)$ , therefore $b<a$ . So let $a = b + c$ where $c$ is a positive integer.

After substituting we get $(b+c)(b+c+2)=b(b+9)$ , which we can rewrite as $b=\frac{-c^2-2c}{2c-7}$ .

A simple inspection reveals that for $c\ge 4$ we'll get a negative value for $b$ and values for $c$ of 1, 2, and 3 evaluate to $\frac{3}{5}, \frac{8}{3},$ and $15$ respectively. As $b$ must be a positive integer $15$ is it's only valid value. $n=15(15+9)=360$ . Since $360$ is the only integer that is both two-separable and nine-separable, $\boxed{360}$ is also the sum of all positive integers that are both two-separable and nine-separable.

Let the number be a. Now $a = x(x+2)$ where x is an positive integer and a is also equal to $a = y(y+9)$ where y is a positive integer. As a should be two-separable and nine-separable this gives us the equation

$x(x+2)=y(y+9)\implies x^{2} - y^{2} +2x -9y = 0$

which is equivalent to the equation of a hyperbola. Solve the equation for positive x and y. Friends you can also take help of wolfram alpha to calculate these integer roots. The link is given below. :)

http://www.wolframalpha.com/examples/CoordinateGeometry.html

Let $n = (k_{1}-1)(k_{1}+1) = k_{2}(k_{2} - 9)$ for positive integers $k_{1}$ and $k_{2}$ with $k_{1} \geq 2, k_{2} \geq 10$ .

Therefore, $k_{1}^{2} = k_{2}^{2} - 9k_{2} + 1$ <=> $k_{2}^{2} - 9k_{2} + (1-k_{1}^{2}) = 0$ .

It implies that $k_{2} = \frac{9 + \sqrt{4k_{1}^{2} + 77}}{2}$ .

Therefore, $4k_{1}^{2} + 77 = k^{2}$ for a positive integer $k$ . Therefore, $k^2 - 4k_{1}^{2} = (k-2k_{1})(k+2k_{1}) = 77$ . It is easy to get that the only $k_{1} \geq 2$ satisfying the above condition is $k_{1} = 19$ .

Therefore, $n = 19^{2} - 1 = 360$ .

I created an extension of this problem. Check out my problem and solution: separable integers reloaded

let x be the number which satisfies both two separable and nine seperable x=a(a+2)=b(b+9) a2+2a-x=0; a=-2+-sqrt 4+4x
b2+9b-x=0; b=-9+-sqrt 81+4x; remember a ,b, x are positive integers(a>0 b>0 x>0) a=-2+sqrt 4+4x and b=-9+sqrt 81+4x it means 4+4x and 81 +4x should be perfect squares simultaneously assume a+2 =p and b+9 =q q2 - p2 = 77 q+p=77 and q-p =1 solutions are q=39 and p=38 similarly q+p=11 and q-p=7 solutions are q=9 and p=2
therefore 77 can be expressed in two possible ways difference of perfect squares 81+4x=1521(square of 39) x=360; 81+4x=81 x=0;(not possible) therefore [360] is the only number which can be written in two seperable and nine seperable