Mambo No. 5

We have $x(x+5)=(y+1)(y+2)(y+3)(y+4)$ $x(x+5)=(y+1)(y+4)(y+2)(y+3)$ $x(x+5)=(y^2+5y+4)(y^2+5y+6)$ Let $z=y^2+5y+5$ . Now, $x(x+5)=(z-1)(z+1)=z^2-1$ We complete the square for $x$ : $x^2+5x+1=z^2$ $\left(x+\dfrac{5}{2}\right)^2-\dfrac{21}{4}=z^2$ $(2x+5)^2-(2z)^2=21$ Now, we use difference of squares: $(2x+5+2z)(2x+5-2z)=21$ Since $21=3\cdot 7$ , 21 can be split four ways.

Case 1: $21=-21\cdot -1$

The sum of the factors is -22, and so $4x+10=-22$ , and $x=-8$ .

Case 2: $21=-7\cdot -3$

The sum of the factors is -10, and so $4x+10=-10$ , and $x=-5$ .

Case 3: $21=3\cdot 7$

The sum of the factors is 10, and so $4x+10=10$ , and $x=0$ .

Case 4: $21=1\cdot 21$

The sum of the factors is 22, and so $4x+10=22$ , and $x=3$ .

Therefore, we have $x=-8,-5,0,3$ .

If $x=-8$ or $x=3$ , $(y+1)(y+2)(y+3)(y+4)=24$ , and $y=-5,0$ . If $x=-5$ or $x=0$ , $(y+1)(y+2)(y+3)(y+4)=0$ , so $y=-4,-3,-2,-1$ .

Overall, our answer is $2+2+4+4=\boxed{12}$ .

We let $n = y+1$ and note that we can turn the RHS of the given equation into $(n^2+3n+1)^2-1$ through the equality $a(a+1)(a+2)(a+3)+1 = (a^2+3a+1)^2$ . Thus, we are looking for $x$ such that $x(x+5)$ is $1$ less than an odd square number (since $n^2+3n+1$ is always odd if $n\in\mathbb{Z}$ ).

Let $m = n^2+3n+1$ . Then $x^2+5x+1 = m^2$ . Since both $x,m\in\mathbb{Z}$ , $m = x+k$ for some integer $k$ . Thus, $x^2+5x+1 = m^2 = (x+k)^2 =$ $x^2 + 2kx + k^2 \to 5x+1 = 2kx + k^2$ . We then have $x = \frac{k^2-1}{5-2k}$ , which we can turn into $x = \frac{1}{4}(-2k - 5 + \frac{21}{5-2k})$ through polynomial long division. In order for $x$ to be an integer, $5-2k$ must divide $21$ , which gives us that $k=-8,-1,1,2,3,4,6,13$ . These values for $k$ give $x = 3,0,0,3,-8,-5,-5,-8$ , respectively. Note that in each of these pairs $(x,k)$ , $x+k$ is odd, which means that $(x+k)^2 = m^2$ is odd, as desired.

Therefore, we have that $x = -8,-5,0,3 \to x(x+5) = 24, 0, 0, 24$ . Thus, there are two values of $x$ which make the LHS equal to $24$ in the original equation and we can easily see that there are two values of $y$ which make the RHS equal to $24$ (these are $y = -5,0$ ). There are also two values of $x$ which make the LHS equal to $0$ and four values of $y$ which make the RHS equal to $0$ (these are $y = -1,-2,-3,-4$ ). We have verified above that no other value can be attained by the LHS and the RHS simultaneously, and thus, the final answer is $2\cdot2 + 2\cdot4 = \fbox{12}$ .

Useful trick: the product of 4 consecutive integers plus one is a perfect square.This follows from:

$z(z+1)(z+2)(z+3)+1 = (\overbrace{z^2+3z}^{\alpha})(z^2 + 3z + 2)+1=\alpha(\alpha+2)+1 = (\alpha+1)^2$

Hence, let's write $x^2 + 5x + 1 = z^2$ . When x>3, we have:

$x^2 + 4x + 4 < x^2 + 5x + 1 < x^2 + 6x + 9$

By symmetry of the parabola, when x<-8, we have $x(x+5)+1 = x'(x'+5)+1$ for $x' = -5-x > 3$ which redcues to the earlier case. Hence, we only need to consider x=-8, ..., 3. This gives:

Case 1 : x = -8, 3. Thus, $x(x+5) = 24$ and one easily sees that y=0, -5. This gives 4 solutions for (x,y).

Case 2 : x = 0, -5. Thus, $x(x+5) = 0$ and we get y=-1, -2, -3, -4. This gives 8 solutions for (x,y).

First, we observe that putting x=0 and x=-5, we can reduce the LHS into 0. So for each of x=0 and x=-5, we can reduce the RHS into 0 in 4 ways i.e. by making y = -1, -2, -3, -4. So we get 8 doubles.

Next, assuming that the RHS and LHS are both non-zero, we have the RHS as a product of 4 consecutive integers. A clever investigation reveals that the RHS is a multiple of 24 always. So, the LHS must also be a multiple of 24.

Hence, we write the equation as x(x+5) = 24m, where m is any integer. Clever investigations yield that if m = 1, then we get x(x+5) = 24 which is always true for x=3. So we get one more double.

Next, we try to find more such doubles. The discriminant is a perfect square, hence

25 + 96m = p^2 where p is an integer. We re-write this as (p+5)(p-5) = 96m.

Clearly if p+5=96 and p-5 = m OR p+5=m and p-5 = 96 we obtain two more doubles. But now let us observe the other values of p+5 and p-5.

Let, p+5 = k and p-5 = 96m/k. Subtracting, we have k - 96m/k = 10.

Now we come to the following conclusion: there must be some k for which the above holds, other than k = m or 96. We use the derivative rule to find that the derivative vanishes for k = 48. So, we have yet another pair.

The derivative rule restricts further pairs.

So answer is 12.

We can write R.H.S. as $\big($ $(y + 1)$ $(y + 4)$ $\big)$ $\big($ $(y + 2)$ $(y + 3)$ $\big)$

i.e. $\big($ $y^{2} + 5y + 4$ $\big)$ $\big($ $y^{2} + 5y + 6$ $\big)$ . Let $y^{2} + 5y =\:$ $z$ $\in\:$ $\mathbb{Z}$ $,$ so the given equation becomes $x(x + 5) =$ $z^{2} + 10z + 24$ ---- $(i)$ .

Now since --- $(i)$ is a quadratic in $x$ having integer roots, so it's discriminant should be a perfect square i.e.

$25 + 4z^{2} + 40z + 96 =\:$ $k^{2}$ for some $k$ $\in\:$ $\mathbb{Z^{+}}$

$\implies\:$ $4z^{2} + 40z + 121 = \:$ $k^{2}$ ------ $(ii)$

Again, since --- $(i)$ is also a quadratic in $z$ having integer roots, so it's discriminant should be a perfect square i.e.

$100 - 96 + 4x^{2} + 20x =\:$ $n^{2}$ for some $n$ $\in\:$ $\mathbb{Z^{+}}$

$\implies\:$ $4(x^{2} + 5x) + 4 =\:$ $n^{2}$

$\implies\:$ $4(z^{2} + 10z + 24) + 4 =\:$ $n^{2}$ $\;\;\;\big( Using (i)\big)$

$\implies\:$ $4z^{2} + 40z + 100 =\:$ $n^{2}$ ------ $(iii)$

$(ii)$ - $(ii)$ gives $,$

$21 =\:$ $k^{2} - n^{2}$

$\implies\:$ $21 =\:$ $(k - n)$ $(k + n)$

Case I : $k - n =\:$ $1$ and $k + n =\:$ $21$ $\implies\:$ $k =\:11$ and $n =\:10$ .So $z =\:0, -10$ $\implies\:$ $y =\:0, -5$ $(Rejecting\;z\;=\;-10\;which\;gives\;non-real\;values\;of\;y)$ and $x =\:-8, 3$ . These give a sub-total of $2$ $\times$ $2 =\:4$ ordered pairs.

Case II : $k - n =\:3$ and $k + n =\:7$ $\implies\:k =\:5$ and $n =\:2$ . So $z =\:-6, -4$ $\implies\:$ $y =\:-1,-2,-3,-4$ and $x =\:0,-5$ giving a sub-total of $2$ $\times$ $4 =\:8$ ordered pairs.

So the total number of ordered pairs of integer solutions is $4 + 8 =\:$ $\boxed{12}$ .

Noticing the symmetry of the roots of the polynomials on each side about $5/2$ inspires the following rearrangement

$4 (2x+5)^2-100 = [(2y+5)^2-9][(2y+5)^2-1]$

and rearranging further

$[2(2x+5)]^2 = [(2y+5)^2-5]^2 + 84.$

Letting $u = 2(2x+5)$ and $v = (2y+5)^2-5$ we can find integers $u,v$ satisfying $u^2-v^2 = (u+v)(u-v) = 84$ by examining the different factorizations of $84$ . So if $84 = ab$ then $2x+5 = (a+b)/4$ so we can ignore factorizations where $a+b \neq 0$ mod $4$ . Thus we only need to consider $2 \times 42, 6 \times 14, -2 \times -42$ and $-14 \times -6$ . Substituting these in allows us to see that the only possible $(x,y)$ belong to the set $\{(3,0), (-8,0), (3,-5), (-8,-5), (0,-1), (0,-2),$ $(0,-3), (0,-4), (-5,-1), (-5,-2), (-5,-3), (-5,-4)\}$ , a total of $\boxed{12}$ solutions.

Add $1$ to both sides and we get $$x^2 + 5x + 1 = (y^2 + 5y + 5)^2$$ For simplicity let $t = y^2 + 5y + 5 \in \mathbb{Z}$ , then $$x^2 + 5x + 1 = t^2$$ 1) For $x \geq 3$ we have $$(x + 2)^2 \leq x^2 + 5x + 1 = t^2 \leq (x + 3)^2$$ which is a contradiction, because perfect square can't be between two consecutive perfect squares.

2) For $x \leq 8$ we have $$(x + 3)^2 \leq x^2 + 5x + 1 = t^2 \leq (x + 2)^2$$ which is a contradiction for the same reason. Hence it suffices to consider finitely many cases (which is very easy thanks to our first equation) $$x = -7,-6,\dots,2$$ to get only $$\boxed{12}$$ solutions.

Let $z=y^2+5y+5$ then the equation becomes $x^2+5x=[(y+1)(y+4)][(y+2)(y+3)]=(z-1)(z+1)=z^2-1$ making a quadratic in terms of $x$ we have: $x^2+5x-(z^2-1)=0$ in order for $x$ to be an integer discriminant of quadratic must be a perfect square: $4z^2+21=w^2$ $(w-2z)(w+2z)=(3\times 7) \ \text{or} \ (-3 \times -7) \ \text{or} \ (1 \times 21) \ ...$ which gives us $z=\pm1,5$ . if $z=\pm1$ then $\small x=0,-5$ and $\small y=-1,-2,-3,-4$

so $8$ orderd pairs from here.if $z=5$ then $y=0,-5$ and $x=8,-3$ so $4$

orderd pair from here. total orderd pairs of $\small (x,y)$ that satisfies the equation are $12$ .

Both expressions in x and y are symmetric. Substitute x=(u-5)/2 and y=(v-5)/2. This gives 4 (u^2 - 25)=(v^2 -1)(v^2 -9).

Then rewrite into form: v^4 + b.v^2 + c. Now it follows: v^2=5 +/- sqrt (u^2 -21). Only solutions for u^2-21 being a perfect square are u^2 equal to 25 or 121. Proof: difference between to perfect squares is at least (n+1)^2 - n^2 = 2n+1 for any n. So u^2-k^2 > 21 for any k as soon as u> 11. The rest is just checking cases and rewriting u and v back to x and y.

Letting $z = y^2 + 5y +5$ , we have:

$\begin{aligned} x^2 + 5x &= x(x+5) \\ &= (y+1)(y+2)(y+3)(y+4) \\ &= [(y+1)(y+4)][(y+2)(y+3)] \\ &= (y^2 + 5y + 4)(y^2 + 5y + 6) \\ &= (z - 1)(z + 1) \\ &= z^2 - 1 \end{aligned}$

So $x^2 + 5x + (1-z^2) = 0$ . Considering this as a quadratic equation in $x$ , we have $x = \frac{-5 \pm \sqrt{5^2-4(1-z^2)}}{2} = \frac{-5 \pm \sqrt{21+4z^2}}{2}$ . Since $x$ is an integer, the discriminant is a square - so we can write $d^2 = 21+4z^2$ , where we may take $d$ to be non-negative. This means $21 = d^2-4z^2 = (d+2z)(d-2z)$ . The only possible (unordered) pairs are $\{d-2z,d+2z\} = \{1,21\},\{3,7\},\{-1,-21\},\{-3,-7\}$ . We want $(d-2z)+(d+2z)=2d$ to be non-negative, so we only take the first two pairs. So we either have: $d=11$ and $z=\pm 5$ , or $d=5$ and $z=\pm 1$ . For $d=5$ , we can calculate $x$ to be $-5$ or $0$ , so $x(x+5) = 0$ . Thus, in both cases, we have $(y+1)(y+2)(y+3)(y+4) = 0$ , so $y=-1,-2,-3, -4$ are the only solutions. For $d=11$ , we can calculate $x$ to be $3$ or $-8$ , so $x(x+5) = 24$ . If $z=5$ , we have $5 = y^2+5y+5$ , so $y(y+5)=0 \Rightarrow y=0 \textrm{ or } -5$ . If $z=-5$ , we have $-5 = y^2+5y+5$ , so $y^2+5y+10=0$ . This has a negative discriminant, so there are no real solutions for $y$ .

So all in all, the solutions for $(x,y)$ are:

$(-5,-1),(-5,-2),(-5,-3),(-5,-4), \\ (0,-1),(0,-2),(0,-3),(0,-4), \\ (3,0),(3,-5),(-8,0),(-8,-5)$

Our answer is thus $\boxed{12}$ .

The equation above is equivalent with $x^2+5x=((y+\frac{5}{2})^2-\frac{1}{4})((y+\frac{5}{2})^2-\frac{9}{4})$ So that $x^2+5x=(y+\frac{5}{2})^4-\frac{10}{4}(y+\frac{5}{2})^2+\frac{9}{16}$ By multiplying the equation with $16$ , we get that $16x^2+80x=(2y+5)^4-10(2y+5)^2+9$ Let $(2y+5)^2=k$ , then $16x^2+80x=k^2-10k+9$ By completing the square, we get $(4x+10)^2-(k-5)^2=84$ By factorizing, we get $(4x+k+5)(4x-k+15)=84$ Consider that $(4x+k+5)$ and $(4x-k+15)$ is the factor of $84$ . By checking all of possible values, we get $6$ pairs $(x,k)$ which their values are $(0,1);(0,9);(3,25);(-5,1);(-5,9);(-8,25)$ . Substituting the value of $k$ to equation $(2y+5)^2=k$ , we get $12$ ordered pairs $(x,y)$ . They are $(0,-1);(0,-2);(0,-3);(0,-4);(-5,-1);(-5,-2);(-5,-3);(-5,-4);(3,0);(3,-5);(-8,0);(-8,-5)$ . So, the answer is $12$ .

Since that $4(y+1)(y+2)(y+3)(y+4)+4=4[(y+1)(y+4)][(y+2)(y+3)]+4$ $=4(y^2+5y+5-1)(y^2+5y+5+1)+4=$ $4(y^2+5y+5)^2-4+4=[2(y^2+5y+5)]^2$ ,

$4x(x+5)+4$ is a perfect square, $4x(x+5)+4=p^2, p=2|y^2+5y+5|\in\mathbb N$

Notice that $p^2=4x(x+5)+4=4x^2+20x+4=(2x+5)^2-21=M^2-21$ , were $M=2x+5$ , $M$ is odd.

So we have $M^2-21=p^2\Rightarrow (*)\quad (M+p)(M-p)=M^2-p^2=21=7*3$ and since that $p\in\mathbb N$ , we have $M+p\ge M-p$ and by the $(*)$ we have four cases:

$M+p=21, M-p=1$ , where $M=11, x=3$

$M+p=7, M-p=3$ , where $M=5, x=0$

$M+p=-1, M-p=-21$ , where $M=-11, x=-8$

$M+p=-3, M-p=-7$ , where $M=-5, x=-5$

If $x=0, -5$ , we have $x(x+5)=0$ , where $(y+1)(y+2)(y+3)(y+4)=0\Rightarrow y=-1, -2, -3, -4$ .

And we have $8$ solutions $(x, y), x=0, -5, y=-1, -2, -3, -4$ .

If $x=-3, 8$ , we have $4x(x+5)+4=100$ and $p=10$ , where $2|y^2+5y+5|=\pm 10\Rightarrow y^2+5y+5=\pm 5\Rightarrow y=-5, 0$ .

So we have more $4$ solutions: $(i, j), x=-3, 8, y=-5, 0$ , totalizing twelve solutions.

Let $s = x+3$ and $z = y+1$

It then becomes $\left(s-3\right)\left(s+2\right) = z\left(z+1\right)\left(z+2\right)\left(z+3\right)$

$s^2-s-6 = \left(z^2+3z\right)\left(z^2+3z+2\right)$

If we let $t = z^2+3z+1$ , it then becomes $s^2 - s - 6 = t^2 -1$

$\left(s-\frac{1}{2}\right)^2 -\frac {25}{4} = t^2 -1$

$\left(s-\frac{1}{2}\right)^2-t^2 = \frac{21}{4}$

$\left(2s-1\right)^2 - \left(2t\right)^2 = 21$

$\left(2s-2t-1\right)\left(2t+2s-1\right) = 21$

$\left\{x,y\right\} \subset \mathbb{Z} \Rightarrow \left\{s,t\right\} \subset \mathbb{Z}$

So we can say that $2s-2t-1 \in \left\{-21,-7,-3,-1,1,3,7,21\right\}$ whilst $2t+2s-1$ is the corresponding factor of $21$ .

Solving we get the set of solutions :

$\left(s,t\right) \in \left\{\left(6,5\right),\left(3,1\right),\left(3,-1\right),\left(6,-5\right),\left(-5,5\right),\left(-2,1\right),\left(0,-1\right),\left(-5,-5\right)\right\}$

Remembering $t = z^2 + 3z + 1$ only $t = -5$ gives no real solutions of $z$ .

Possible solutions of $\left(x,y\right)$ are therefore $\left(3,0\right),\left(3,-5\right),\left(0,-1\right),\left(0,-4\right),\left(0,-2\right), \left(0,-3\right),\left(-8,-5\right), \left(-8,0\right), \left(-5,-1\right), \left(-5,-4\right), \left(,-3,-2\right), \left(-3,-3\right)$

Therefore there are 12 ordered pairs of integers $x$ and $y$ that satisfy the given equation.

Case 1: $x(x+5)=0$ . Then $x\in\lbrace-5,0\rbrace, y\in\lbrace-4,-3,-2,-1\rbrace$ .

Case 2: $x(x+5)\ne 0$ .Group the factors $(y+1)(y+4) = y^2+5y+4$ and $(y+2)(y+3)=y^2+5y+6$ .Let $z=y^2+5y+4$ . So, $x(x+5) = z(z+2)$ . Denote this product as N. We rule out the possibility that N is negative since that would only be possible if $z=-1$ but then $x(x+5)=-1$ .

So, N is positive, and can be written as a product of numbers which differ by 5, and by 2. These factors can be both positive, or both negative. If these are both positive, then one is greater than $\sqrt{N}$ and one is less than $\sqrt{N}$ . If $0 < z\le x$ then $z(z+2) \le x(x+5)$ . If $0< x \le z+2$ then $z(z+2) \ge (x+2)(x+4) = x^2+6x+8 > x(x+5)$ . So, the only possibility is that $z=x+1$ and then $x(x+5)=x^2+5x = (x+1)(x+3) = x^2+4x+3$ so $x=3, N=24=3\times 8=4\times 6$ . If $y^2+5y+4=4$ then $y=-5,0$ so $(x,y)=(3,-5)$ or $(3,0)$ .

The negative factorizations are symmetric, and they also come from $N=24=(-8)(-3)=(-6)(-4)$ : $(x,y)=(-8,-5)$ or $(-8,0)$ .

First of all, there are some trivial solutions where one of the factors is zero, I hope I can give these without further explanation:

(0,-1) (0,-2) (0,-3) (0,-4) (-5,-1) (-5,-2) (-5,-3) (-5,-4)

For the non-trivial solutions, we can simplify the equality slightly:

x(x+5) = (y+1)(y+4) . (y+2)(y+3)
       = z(z+2)     where z=(y+1)(y+4)

Now, suppose there is some solution involving z = a (a > 0).

Then z = -(a+2) is also a solution:

z(z+2) = -(a+2)(-(a+2)+2) 
       = -(a+2)(-a)
       = a(a+2)

and -(a+2) must be negative.

So every solution for which (z > 0) must have a corresponding solution for which (z < 0), and x is the same.

Similar logic shows that every solution for which (x > 0) has a corresponding solution for which (x < 0), and z is the same.

So, we can restrict our search to cases where (x > 0) and (z > 0). Because any such solution gives the following set of solutions:

(x, z)  (-(x+5), z)   (x, -(z+2))   (-(x+5), -(z+2))

Looking at the equality again:

x(x+5) = z(z+2)         where x > 0, z > 0

We can write: z = x + h where h > 0 , giving:

 x(x+5) = (x+h)(x+h+2)
     5x = 2x + 2xh + h(h+2)
     3x = 2xh + h(h+2)
(3-2h)x = h(h+2)

Since h and x are positive, (3 - 2h) must be positive too, meaning that (h = 1) is the only candidate solution.

Solving with h=1 gives x=3, z=4; and z=4 corresponds to y=0 or y=-5.

So we have exactly four non-trivial solutions:

(3, 0)  (3, -5)  (-8, 0)  (-8, -5)

for a total of 12 solutions.