Ahmad's equation

We have $p = \frac{x^2+xy-2y^2}{30} = \frac{(x-y)(x+2y)}{30}$ Now, since $x-y$ and $x+2y$ are congruent modulo 3, and 30 divides their product, both $x-y$ and $x+2y$ are multiples of 3. Let $x-y = 3k, x+2y = 3h$ . But then $p = 3\frac{hk}{10} \quad\Longrightarrow\quad 3|p \quad\Longrightarrow\quad p = 3\text{ and }hk=10$ Since for every ordered pair $(h,k)$ we find exactly one ordered pair $(x,y) = (2k+h,h-k)$ with an injective map, we just need to count the number of distinct ordered pairs $(h,k)$ of integers with the property that $hk=10$ . This number is $2\cdot\tau(10) = \boxed{8}$ .

We expand, and write as a quadratic in $y$ : $2y^2-xy+(30p-x^2)=0$ Using the quadratic formula, $y=\dfrac{x\pm\sqrt{9x^2-240p}}{4}$ Now, note that $9x^2-240p=3(3x^2-80p)$ must be a square. Therefore, since 3 divides it, 9 must too, and so $p=3$ . $y=\dfrac{x\pm 3\sqrt{x^2-80}}{4}$ Let $x^2-80=k^2$ . We have that $(x-k)(x+k)=80$ We know that $80=1\cdot 80=2\cdot 40=4\cdot 20=5\cdot 16=8\cdot 10$ , and also we know that the sum of $x-k+x+k=2x$ is even, and so we can divide 80 into 2 and 40, 4 and 20, or 8 and 10. From this we get that $x=\pm 9,\pm 12,\pm 21$ . It is easy to find which values of $x$ give integer values of $y$ , and our solutions are:

$(-21,9)$ , $(-12,-9)$ , $(-12,3)$ , $(-9,-3)$ , $(9,3)$ , $(12,-3)$ , $(12,9)$ , $(21,-9)$

The answer is $\boxed{8}$ .

Expanding and rearranging gives the equation in the form

$\qquad \qquad 30p = x^2+xy-2y^2$

which can be factorised as

$\qquad \qquad 30p = (x+2y)(x-y)\\$

Now, either $(x+2y)$ or $(x-y)$ must have a factor of $3$ , since there is a factor of $3$ in the LHS; however:

$\qquad \qquad (x+2y) = (x-y) + 3y$

So if one of them is divisible by $3$ , then so is the other, since they differ by a multiple of $3$ . This means that there must be an extra multiple of $3$ on the LHS. The only way this can happen is if $p=3$ - no other value of $p$ is possible.

Then we have:

$\qquad \qquad 90 = (2y+x)(x-y)$

Each of the brackets has exactly one factor of $3$ , so subject to that restriction there are 8 ways to split $90$ into the two factors $(x+2y),(x-y)$ -

$\qquad \qquad (\pm 3, \pm 30)\\ \qquad \qquad (\pm 30, \pm 3)\\ \qquad \qquad (\pm 6, \pm 15)\\ \qquad \qquad (\pm 15, \pm 6)\\$

These each give the only unique values of $x,y$ that work. At this point we can say, after verifying that every pair above gives integer solutions, that there are $8$ possible pairs $x,y$ , which are:

$\qquad \qquad (21,-9),(-21,9),\\ \qquad \qquad (12,9),(-12,-9),\\ \qquad \qquad (12,-3),(-12,3), \\ \qquad \qquad(9,3),(-9,-3)$

Quick,what makes a prime number different from others?The fact that it always has only 2 positive divisors!So,what if,what if,we can turn our solution into something of the form

$ab=p$ where $a,b$ are integers and $p$ is a prime.

Then,if $a$ and $b$ are integers,then we have only 8 possible ordered pairs,counting negatives of course.So we want to keep $p$ on the right side only.Now let's look at the problem at hand.

$x(x+y)=2(y^2+15p)$

We expand,and keep $p$ on the left side.

$x^2+xy+2y^2=30p$

and now we try to factorize the left side.We try completing the square.

$x^2-2xy+y^2-3y^2+3xy=30p\implies (x-y)^2+3y(x-y)=30p\implies (x-y)(x+2y)=30p$

Ha!I know this isn't exactly the same as what we wanted.But it's good enough.Now,suppose that $x-y=m,x+2y=n$ where m and n are integers.Solving the equation gives us $y=\frac{n-m}{3}$ .But since $y$ must be an integer,, $n-m$ i.e. difference of the two factors must be divisible by 3.Now,we come back to our original equation.

$(x-y)(x+2y)=30p$

We sort by Values of $x-y$ and $x+2y$ .But recall that the difference of the factors must be divisible by 3.For example,if $x-y=\pm 2,x+2y=\pm 15p$ ,then there is no prime p such that the difference of the factors is divisible by 3.So this lets us eliminate some possibilities.Now we do the tedious part.We are going to count unordered pairs,not ordered pairs.

$1)(\pm 30,\pm p),2)(\pm 2p,\pm 15),3)(\pm 6,\pm 5p)$

For each of these,only $p=3$ works,which is obvious.However the pairs (2) and (3) are equivalent,after putting in $p=3$ since they yield the same equations.Now,we are left with 2 values. sighs

$(x-y,x+2y)=(\pm 30,\pm p),(\pm 2p,\pm 15)$ and

$(x+2y,x-y)=(\pm 30,\pm p),(\pm 2p,\pm 15)$

This gives us 8 solutions.

x(x + y) = 2(y^2 + 15p) => 30p = x^2 + xy - 2(y^2) = (x - y)(x + 2y) => (x - y)(x + 2y) = 2 * 3 * 5 * p

Let (x - y) = a and (x + 2y) = b where a and b are also integers. We get x = (2a + b)/3 and y = (b - a)/3.

As ab = 2 * 3 * 5 * p, one of a and b would be divisible by 3. If a is divisible by 3, we get x = 2k1 + (b/3) and y = (b/3) - k1. If b is divisible by 3, we get x = {(2a)/3} + k2 and y = k2 - (a/3).

Thus, we find that b in the first case and a in the second case also have to be divisible by 3 for x and y to be integers. This can only happen when the prime p is nothing but 3.

Therefore, we get (x - y)(x + 2y) = 2 * 3 * 5 * p = 2 * 3 * 5 * 3 = 90. 90 = 1 * 90 = 2 * 45 = 3 * 30 = 5 * 18 = 6 * 15 = 9 * 10

The equations we need to solve to determine integral values of x and y are: (x - y) = {-30, - 15, -6, - 3, 3, 6, 15, 30} with the corresponding (x + 2y) = {-3, - 6, - 15, - 30, 30, 15, 6, 3} So, there are 8 ordered set of integers.

Solving for all the possible pairs of equations, we get (x, y) = {(-21, 9), (-12, 3), (-9, -3), (-12, -9), (12, 9), (9, 3), (12, -3), (21, -9)}.

We have:

$x( x + y ) = 2( y^{2} + 15p ) <=> x^{2} - y^{2} + xy - y^{2} = 30p <=> ( x - y )( x + y ) + x( x - y ) = 30p <=> ( x - y )( x + 2y ) = 30p (1)$

So /(30p/) is the product of $(x-y)$ and $(x+2y)$ . Let $a$ and $b$ be positive integers such that $a \times b = 30p$ . From (1) we get two equations:

$x + 2y = b$

$x - y = a$

$(2)$

The difference gives:

$-3y = a - b <=> a+3y=b <=> 3( \frac{a}{3} + y)=b$

Because b is an integer $\frac{a}{3} + y$ must be an integer too therefore 3 divides both a and b. Now (2) has exactly one solution (x,y) so the number of possible values a or b can take is the how many ordered sets of integers (x,y) satisfy the parameters given in the problem. We List the possibilities $(a,b): (3,10p),(6,5p),(15,2p),(30,p),(p,30),(2p,15),(5p,6),(10p,3)$ . Hence the answer is 8.

Rearranging the equation as $x^2 - y^2 + xy - y^2 = 30p$ , we see the LHS factors to $(x-y)(x+2y)$ . The difference between $x+2y$ and $x-y$ is $3y$ . Since the RHS is a multiple of 3, either $x+2y$ or $x-y$ must be a multiple of 3, but since their difference is $3y$ , we see that both $x+2y$ AND $x-y$ have to be multiples of 3, so $30p$ must be a multiple of 9, and the only prime $p$ that satisfies this is $p=3$ .

We are now seeking how many ordered sets of integers $(x,y)$ that satisfy: $(x-y)(x+2y)=90$ . 90 can be factored into two multiples of 3 in the following ways: 3 * 30, 6 * 15. However, note that we can have both factors be negative, and we also have 2 ways to assign each factor to being either $x+2y$ or $x-y$ . Thus, there are 2 * 2 * 2 = 8 total ordered pairs that work.

We can rearrange Ahmad's equation to obtain $30p = (x - y)(x + 2y)$ . Setting $a = x - y$ yields $30p = a (a + 3y)$ , whence $30p = a^2 + 3ay$ . (1)

Since the left hand side of this equality is divisible by 3, so is the right hand side, and so $3 | a^2$ , which shows that $3 | a$ . In other words, $a = 3k$ for some integer $k$ . By substitution on (1), we get $30p = 9k^2 + 9ky$ . (2)

We can therefore conclude that $9 | 30p$ , and since $p$ is a prime, $p = 3$ . Replacing and simplifying from (2), we get $10 = k^2+ ky$ or, equivalently, $10 = k (k + y)$ .

Finding the solutions of the problem now ammounts to studying all possible factorizations of 10 (which are 8, if we allow negative factors). Each one will yield a distinct solution to Ahmad's equation.

The quadratic formula gives us $x = \frac{1}{2} \times \left( -y \pm \sqrt{3(40p + 3y^2)} \right)$ .

Since $x$ is an integer, $3(40p + 3y^2)$ must be a perfect square. This can only be true if $40p + 3y^2$ has a factor of 3. Suppose $p \neq 3$ . Then $p$ must be congruent to either 1 or 2 modulo 3 (as otherwise it would be divisible by 3, and either be 3, or a non-prime). Now $3y^2$ is obviously congruent to zero mod 3, and thus the entire expression is congruent to either 1 or 2 mod 3, i.e. does not have 3 as a factor. But then $3(40p + 3y^2)$ cannot be a perfect square. Thus by contradiction, we have $p = 3$ .

By substitution we have $x = \frac{1}{2} \times \left( -y \pm 3 \sqrt{40 + y^2} \right)$ . Now $40 + y^2$ is only a perfect square for $y = \pm 3$ and $y = \pm9$ - this may be verified with casework, or by observing that the difference between two adjacent squares follows the odd numbers. Thus obtaining a difference of $40$ by adding up consecutive odd numbers can only be done by adding $7 + 9 + 11 + 13$ or $19 + 21$ , and these are the differences between $3^2$ and $7^2$ , and $9^2$ and $11^2$ respectively.

Hence we have four values of $y$ . Each value of $y$ clearly gives two solutions for $x$ , and these values will all be integers. This is apparent as $y$ is odd in all cases, and $3 \times \sqrt{40 + y^2}$ is the product of two odd numbers and thus odd - hence, adding them together or subtracting one from another will always give an even number. Thus, the multiplication by $\frac{1}{2}$ is safe and will leave all of the solutions for $x$ safely as integers. (Admittedly, it is probably easier to simply verify this with a substitution.)

Thus the answer is $4 \times 2 = \boxed{8}$ .

We have $x(x+y)=2(y^2+15p)\Rightarrow x^2+xy-2y^2=30p\quad (*)\Leftrightarrow$

$(x+2y)(x-y)=30p\Rightarrow 3|x+2y\quad$ or $\quad 3|x-y\Rightarrow 3|x+2y, x-y$ since that $x-y\equiv x+2y\pmod{3}$

where $9|(x-y)(x+2y)\Rightarrow 9|30p\Rightarrow 3|10p$ $\Rightarrow 3|p\Rightarrow p=3$ since that $p$ is a prime number and $mdc(10, 3)=1$

Now, by $(*)$ and $p=3|x+2y, x-y$ we have $\frac{x+2y}{3}*\frac{x-y}{3}=10\Leftrightarrow (**)\quad f*g=10$ , $f=\frac{x+2y}{3}, g=\frac{x-y}{3}$ integers and explicitly $x=2g+f, y=f-g$ .

Then every solution of $(**)$ is a solution of $(*)$ and contrariwise.

Thus and the number of solutions of $(**)$ is the number of integers divisors of $10$ . How $d(10)=8$ , we have $8$ solutions.

The given equation is equivalent to $x^2+xy-2y^2=(x+2y)(x-y) = 30p = ab,$ where $a=x+2y$ and $b=x-y$ . This implies $3\mid ab$ and since $3$ is prime, it follows that $3 \mid a$ or $3\mid b$ . Assume that $3 \mid a$ . Since $a-b = 3y$ , we have $3 \mid (b-a).$ Hence $3 \mid ((b-a)+a)$ or $3 \mid b.$ Similarly, if $3\mid b$ , then $3 \mid a.$ Since $30p$ factors into primes as $2\cdot 3 \cdot 5 \cdot p$ if $p\neq 3,$ then $3$ would be a factor of either $a$ or $b$ but not both, contradicting to the previous argument we have just made. Therefore, $p = 3$ and $3$ is a common factor of $a$ and $b.$ This implies we have only $4$ possible set equations, namely $\{x+2y,x-y\} = \{6, 15\}, \{-6,-15\}, \{3, 30\}, \,\text{or}\, \{-3,-30\}.$ Each set equation gives two possible systems of equations, each producing an integer solution $(x,y).$ Hence there is a total of $4\times 2 = 8$ pairs $(x,y)$ of integers satisfying the given equation.

Expand and rearrange to get: $30p = x^2+xy-2y^2 = (x-y)(x+2y)$

Let $a=x-y,b=x+2y$ . Then $ab=30p$ and $2a+b=3x$ (or $b-a=3y$ ). The first equality implies that at least one of $a,b$ is divisible by 3. The second equality then restricts that both $a$ and $b$ are divisible by 3, for $x$ (similar, $y$ ) to be an integer. Then the first equality again implies that $p=3$ .

So $ab=90,3|a,3|b$ . There are 8 choices:

1. $a=3,b=30$ then $x=-6, y=9$
2. $a=6,b=15$ then $x=3, y=3$
3. $a=15,b=6$ then $x=12, y=-3$
4. $a=30,b=3$ then $x=21, y=-9$

and the negative versions (that is, $a'=-a,b'=-b,x'=-x,y'=-y$ ) of each of the four cases above.

Therefore, there are 8 choices for $(x,y)$ that work.

solve for the left and right sides thus,

$x^{2}$ + xy = $2y^{2}$ + 30p

then,

$x^{2}$ + xy - $2y^{2}$ = 30p

using factoring,

(x + 2y)(x - y) = 30p

now, we know that the two solutions of x and y on both factors and disregarding the 'p".. thus,

(x + 2y) or (x - y)

x = 1 and y = 30 or x = 2 and y = 15 or x = 3 and y = 10 or x = 5 and y = 6 or

x = 6 and y = 5 or x = 10 and y = 3 or x = 30 and y = 1

that are 8 values

Equation is equivalent to (x+2 y) (x-y)=30 p.
Some of left side multiplicants should divide by 3, but then the other divedes too, so 30
p divedes by 9, so p = 3.
Denote x-y=3 z. Then z (z+y)=10, z may be any divisor of 10 (there are eight of them), y=10/z-z, x=10/z+2*z.