Michael's integer solutions

Multiply out the equation:

$a^2(a^2 - 2) = b^2 + 7$

$a^4 - 2a^2 = b^2 + 7$

Bring a and b terms on one side:

$a^4 - 2a^2 - b^2 = 7$

Make a perfect square out of a terms by adding 1 on both sides:

$(a^4 - 2a^2 + 1) - b^2 = (7 + 1)$

$(a^2 - 1)^2 - b^2 = 8$

Use difference of squares to get the two factors:

$(a^2 - 1 - b)(a^2 - 1 + b) = 8$

Try all cases of the factors of 8: Case 1: $1 \times 8$ AND $8 \times 1$

$a^2 - b - 1 = 1$ OR $a^2 - b - 1 = 8$

$a^2 +b - 1 = 8$ OR $a^2 + b - 1 = 1$

By adding both terms:

$2a^2 - 2 = 9$ -> a is not an integer; case does not work

Case 2: $2 \times 4$ AND $4 \times 2$

$a^2 - b - 1 = 2$ OR $a^2 - b - 1 = 4$

$a^2 +b - 1 = 4$ OR $a^2 + b - 1 = 2$

By adding both terms:

$2a^2 - 2 = 6$

$a = \pm 2$

Substitute $\pm 2$ back into both possible equations:

$4 - b - 1 = 2$

$b = 1$

$4 - b - 1 = 4$

$b = -1$

Coordinates are $(\pm 2, \pm 1)$ for a total of $4$ ordered pairs possible

$a^2$ ( $a^2$ -2 ) = $b^2$ + 7

$a^4$ - 2 $a^2$ = $b^2$ + 7

$a^4$ - 2 $a^2$ + 1 -1 = $b^2$ + 7

$(a^2 -1)^2$ - $b^2$ = 8

( $a^2$ + $b^2$ - 1) ( $a^2$ - $b^2$ - 1) = 8

As a and b are integers, $a^2$ - $b^2$ - 1 and $a^2$ + $b^2$ - 1 are integers

Factors of 8: 1,2,4,8

When ( $a^2$ + $b^2$ - 1) and ( $a^2$ - $b^2$ - 1) = 1 and 8, $a^2$ = $\frac {11}{2}$ and a isn't an integer

When ( $a^2$ + $b^2$ - 1) and ( $a^2$ + $b^2$ - 1) = 2 and 4

( $a^2$ + $b^2$ - 1) = 2, ( $a^2$ - $b^2$ - 1) = 4

2 $a^2$ - 2 = 6

$a^2$ = 4

a = -2, 2 and b = -1, 1

Therefore, there are 4 ordered pairs: (-2,-1), (-2,1), (2,-1), (2,1)

We are given $a^4-2\cdot\,a^2-(b^2+7)\,=0$ --- $(i)$

Since $a\in\mathbb{Z}$ so $a^2\in\mathbb{Z^+}$ .

So discriminant of $(i)$ is perfect square, i.e.

$4+4\cdot\,(b^2+7)=k^2$ for some $k\in\mathbb{Z^+}$ and $k>b$

$\rightarrow\,(k-2b)(k+2b)=32$ --- $(ii)$

Thus $a^2=\frac{2\pm\,k}{2}$ $\rightarrow$ $k$ is even.

From $(ii)$ we get $(k-2b,k+2b)=(4,8)$ as the only ordered pair $\rightarrow$ $k\,=6$

Now plugging $k=6$ in $(i)$ we find $a=\pm\,2$ and plugging $a=\pm\,2$ in our given equation we get $b=\pm\,1$ .

So the ordered pairs $(a,b)$ are $(\pm\,2,\pm\,1)$ which gives a total of $\boxed{4}$ ordered pairs.

Simplifying, $a^4-2a^2=b^2+7$ . Adding $1$ to both sides, $a^4-2a^2+1=b^2+8$ , so $(a^2-1)^2=b^2+8$ .

One square is $8$ more than another square. Either the difference between $k^2$ and $(k+1)^2$ or the difference between $k^2$ and $(k+2)^2$ is $8$ for $k\in\mathbb{N}$ . The first case is impossible because the difference between consecutive square numbers is always odd. The only case where this is possible is the second case.

$(k+2)^2-k^2=8 \Rightarrow 4k+4=8 \Rightarrow k=|b|=1 \Rightarrow b=±1$ .

$(a^2-1)^2=9 \Rightarrow a^2-1=±3 \Rightarrow a^2=4 \Rightarrow a=±2$ .

Thus, the ordered pairs $(a,b)$ that are solutions to the problem are $(±2,±1)$

Expanding, we have $a^4-2a^2=b^2+7$ . Since the LHS looks like a perfect square, we add 1 to both sides to complete the square: $(a^2-1)^2=b^2+8$ . Moving the $b^2$ to the left side to create a difference of squares, we have $(a^2-1)^2-b^2=8 \Rightarrow (a^2+b-1)(a^2-b-1)=8$ .

Since $a, b$ are integers, $a^2+b-1, a^2-b-1$ are integers.

Note that if $a^2+b-1=m, a^2-b-1=n$ , then $a^2+b=m+1, a^2-b=n+1 \Rightarrow a^2=\frac{m+n+2}{2}$ . By the trivial inequality, $a^2 \ge 0$ , so $\frac{m+n+2}{2} \ge 0 \Rightarrow m+n \ge -2$ . There is an integral solution for $a$ if and only if $\frac{m+n+2}{2}$ is a perfect square.

Since integers $m, n$ satisfy $mn=8$ and $m+n \ge -2$ , we have the following cases:

$m=1, n=8: \frac{m+n+2}{2}=\frac{11}{2}$ . Since this is not a perfect square, no solutions exist

$m=8, n=1: \frac{m+n+2}{2}=\frac{11}{2}$ : No solutions.

$m=2, n=4: \frac{m+n+2}{2}=4 \Rightarrow a= \pm 2$ . In this case, $a= \pm 2 \Rightarrow 4+b-1=2 \Rightarrow b= -1$ . This makes a total of 2 solutions.

$m=4, n=2: \frac{m+n+2}{2}=4 \Rightarrow a= \pm 2$ . In this case, $a= \pm 2 \Rightarrow 4+b-1=4 \Rightarrow b=1$ , giving us 2 more solutions.

In total, we have FOUR=4 Solutions.

Let $a^2 = x$ , then $x^2 - 2x - 7 = b^2$ , i.e. $x = \sqrt{b^2 + 8} \pm 1$ hence there exists $c \in \mathbb{Z}$ such that $b^2 + 8 = c^2 \Leftrightarrow (c - b)(c + b) = 8$ Now we consider eight cases $(c - b,c + b) \in \{(-8,-1),(-4,-2),(-2,-4),(-1,-8),(1,8),(2,4),(4,2),(8,1)\}$ to find $4$ solutions $(b,c) = (-1, \pm 3)$ and $(b,c) = (1,\pm 3)$ . From this we get $4$ solutions $(b,x) = (\pm 1,-2$ ) and $(b,x) = (\pm1, 4)$ , the first two solutions are extraneous because $x \geq 0$ , hence we get four solutions $(a,b) = (\pm 2,1)$ , $(a,b) = (\pm 2, - 1)$ . Hence the answer is $$\boxed{4}$$

Notice that if $(a, b)$ is a solution, then $(a, -b), (-a, b), (-a, -b)$ are solutions as well.

$a^2(a^2 - 1) = b^2 + 7 \implies a^4 - 2a^2 - (7 + b^2) = 0$

From the quadratic formula,

$a^2 = \frac{2 \pm \sqrt{4 + 4(7 + b^2)}}{2} = 1 \pm \sqrt{8 + b^2}$

Therefore, $8 + b^2$ is a perfect square.

If $b > 5, b^2 > 25, (b+1)^2 > 36 \implies (b+1)^2 - b^2 > 11 \implies (b+1)^2 > 11 + b^2$

But, $b^2 < b^2 + 8 < b^2 + 11 < (b+1)^2$

This means that $b^2 + 8$ falls between the squares of two consecutive integers. Therefore, $b^2 + 8$ cannot be a perfect square.

Therefore, $b \leq 5$

Similarly, if $b < -5, b^2 > 25, (b+1)^2 > 36\implies (b+1)^2 - b^2 > 11 \implies (b+1)^2 > 11 + b^2$

But, $b^2 < b^2 + 8 < b^2 + 11 < (b+1)^2$

This means that $b^2 + 8$ falls between the squares of two consecutive integers. Therefore, $b^2 + 8$ cannot be a perfect square.

Therefore, $b \geq -5$

Putting in values of $b$ from $-5$ to $5$ , we see that $(2, 1), (-2, 1), (-2, -1), (2, -1)$ are the only solutions. Therefore, there are $\boxed{4}$ solutions.

For convenience let $m = a^2$ and $n = b^2$ we have

$m(m - 2) = n + 7$

$\iff m^2 - 2m - (n+7) = 0$

Solving using the quadratic formula and simplifying we get,

$m = 1 \pm \sqrt{n+8}$

rewriting using original symbols,

$a^2 = 1 \pm \sqrt{b^2 + 8}$

as $a$ is an integer, $a^2$ is a positive integer which happens when $b^2 + 8$ is a perfect square, i.e.,

$b^2 + 8 = p^2$ where p is an integer.

Rearranging we get

$p^2 - b^2 = 8$

$\iff (p - b)(p + b) = 8$

at this point we equate the factors on the left hand side to the factors of 8 on the right hand side as follows:

$(p - b)(p + b) = (8)(1) = (2)(4) = (4)(2) = (8)(1)$

if $p - b = 8$ and $p + b = 1$ , solving for $p$ and $b$ give non integral values. if $p - b = 2$ and $p + b = 4$ , solving for $p$ and $b$ gives integral values. Here we get

$b = 1$ and $p = 3$ .

Substituting this value of $b$ in $a^2 = 1 \pm \sqrt{b^2 + 8}$ we get,

$a^2 = 1 \pm 3$

We take only the plus sign because the square of a number cannot be negative (since we are not dealing with complex numbers here).

$a^2 = 4$ $\iff a = \pm 2$

2 of the ordered pairs are $(2, 1)$ and $(-2, 1)$

Continuing the process in a similar fashion for the cases when

$p - b = 4$ and $p + b = 2$

and the case when

$p - b = 8$ and $p + b = 1$

gives 2 more ordered pairs which are $(2, -1)$ and $(-2, -1)$ .

Therefore there are $\boxed{4}$ ordered pairs to the equation.

a^2 (a^2-2) = b^2 + 7 (a^4-2a^2+1) - b^2 = 8 (a^2-1)^2 - b^2 = 8 DIfference of two squares is 8.. So let us find two no. From 1, 4, 9, 16, 25, 36... whose difference is 8.. there is only one ordered pair present which is ( 9, 1) So b^2 = 1 and (a^2-1)^2 = 9 a= 2 or -2 and b = 1or - 1 So.. 4 ordered pairs

$a^{2}(a^{2} - 2) = b^2 + 7$

$a^{4} - 2a^{2} = b^{2} + 7$

$a^{4} - 2a^{2} + 1 = b^{2} + 8$

$(a^{2} - 1)^{2} = b^{2} + 8$

$(a^{2} - 1)^{2} - b^{2} = 8$

$(a^{2} + b - 1)*(a^{2} - b - 1) = 8$

Solution 1: $a^{2} + b - 1 = 1, a^{2} - b - 1 = 8$

Sum both equations to obtain $2a^{2} - 2 = 9$ , or $2a^{2} = 11$ . $a^{2} = 11/2$ , thus there are no integer solutions for $a$ .

Solution 2: $a^{2} + b - 1 = 2, a^{2} - b - 1 = 4$

Sum both equations to obtain $2a^{2} - 2 = 6$ , or $2a^{2} = 8$ . Then $a^{2} = 4$ , which gives us $a = 2$ and $a = -2$ . For either of these values, we have:
$4 + b - 1 = 2$ , so $b = -1$ and we have the solutions: $(2, - 1), (-2, -1)$

Solution 3: $a^{2} + b - 1 = 4, a^{2} - b - 1 = 2$

Sum both equations to obtain $2a^{2} - 2 = 6$ , or $2a^{2} = 8$ . Then $a^{2} = 4$ , which gives us $a = 2$ and $a = -2$ . For either of these values, we have:

$4 + b - 1 = 4$ , or $b = 1$ . We then have the solutions: $(2, 1), (-2, 1)$

Solution 4: $a^{2} + b - 1 = 8, a^{2} - b - 1 = 1$

Sum both equations to obtain $2a^{2} - 2 = 9$ , or $2a^{2} = 11$ . $a^{2} = 11/2$ , thus there are no integer solutions for $a$ .

Note that if we considered factoring over negative numbers, we would have results of the type $a^2 = k$ , where $k$ is a negative number, and thus there will be no solutions over the integers. Finally, we can see that we have a total of 4 possible solutions for the equation over the integers.