Arshdeep's ordered pairs

$a^2(b^2+1)+b^2(a^2+16)=448\implies 2a^2b^2+a^2+16b^2=448$

At this point,I stopped and thought about it.After some failed attempts at completing the square,I tried the Favorite Factoring Trick and got

$a^2(2b^2+1)+8(2b^2 +0 )=448$

See the $1$ missing beside the $2b^2$ ?What if we add an $8$ to both sides?We can fully factor it to get

$(a^2+8)(2b^2+1)=456=2^33\cdot 19$

The second factor must always be odd,The only odd factors of $456$ are $3,19,57$ ,and $1$ . Plugging in the values yields the following solutions:

$(12,1),(12,-1),(-12,-1),(-12,1),(4,3),(4,-3),(-4,3),(-4,-3)$

Hence,the answer is $8$ .

$a^2(b^2+1)+b^2(a^2+16)=448 \Rightarrow 2b^2=\frac{448-a^2}{(a^2+8)}$ $=-\frac{a^2-448}{a^2+8}=-\frac{(a^2+8)-448-8}{a^2+8}=-1+\frac{456}{a^2+8}=\frac{2^3*3*19}{a^2+8}-1$ . $\frac{2^3*3*19}{a^2+8} \mbox{ must be an odd number , so 8 must cancel with the denominator . }$ $\Rightarrow a^2+8 \in M_{8} \mbox{(the multiples of 8)} \Rightarrow a^2\in M_{8} \Rightarrow$ $a^2 \in \{ 0,16,56,144,256,400\}$ $\mbox{But also } a^2+8 \le 456/2 \Rightarrow a^2 \in \{0,16,56,144\}$ . $\mbox{Checking these values we see that only 16 and 144 work out .} \Rightarrow$

$a=\pm 4 | b=\pm 3$ or $a=\pm 12 | b=\pm 1 \Rightarrow \mbox{There are 8 ordered pairs.}$

(a^2) (b^2 + 1) + (b^2) (a^2 + 16) = 448 => a^2 + 2 (a^2) (b^2) + 16 (b^2) = 448 => 2 (a^2) (b^2) + 16 (b^2) + a^2 + 8 = 448 + 8 = 456 => 2 (b^2) (a^2 + 8) + (a^2 + 8) = 456 => {2 (b^2) + 1} (a^2 + 8) = (2^3) 3 19

Since a & b are integers both {2 (b^2) + 1} and (a^2 + 8) are positive integers with {2 (b^2) + 1} being odd. Therfore, {2*(b^2) + 1} must be equal to one of the odd factors of 456 viz. 1, 3, 19, and 57.

2 (b^2) + 1 = 1 => b^2 = 0 => b = 0; a^2 = 448 => a = +/- 8 sqrt(7) --> No Solution 2 (b^2) + 1 = 3 => b^2 = 1 => b = +1, -1; a^2 = 144 => a = +12, -12 --> 2 2 = 4 ordered pair of solutions 2 (b^2) + 1 = 19 => b^2 = 9 => b = +3, -3; a^2 = 16 => a = +4, -4 --> 2 2 = 4 ordered pair of solutions 2 (b^2) + 1 = 57 => b^2 = 28 => b = +/- 4 sqrt(7) --> No Solution

Therefore, there are 8 solutions. (a, b) = {(-12, -1), (-12, 1), (12, -1), (12, 1), (-4, -3), (-4, 3), (4, -3), (4, 3)}

$2a^2b^2 + a^2 + 16b^2 =448$ $(a^2 + 8)(2b^2 + 1) = 456$

Since $a$ and $b$ are integers, so are the factors on the RHS. Also note that $2b^2 + 1$ is odd, so when we examine the factors of $456 = 2^3 \times 3 \times 19$ we only have to check the cases when $2b^2 + 1$ is $3$ , $19$ or $3 \times 19$ . It follows the solutions over the positive integers are $(a,b) = (12,1), (4,3)$ . Considering negative integers we get 8 solutions in total

Rewrite the LHS, we have 2a^2 b^2 + a^2 + 16 b^2 = 448. Adding 8 to the both sides and manipulating the expression, we get (2b^2 + 1)(a^2 +8) = 456.

Note that 456 = 8 x 3 x 19 Since 2b^2 + 1 is non negative odd, its only possible value are 1, 3, 19, 57. Checking case by case whether a is an integer or not, the only possible pairs (a^2, b^2) are (1, 144) and (9, 16). This yields 8 cases of solutions.

By solving the given equation we find that a^2 = (16(28-b^2))÷(2b^2+1). As L.H.S is positive R.H.S should also be positive this implies that b^2<28 or b<=5. By putting values of b^2 which are 1,4,9,16 and 25. At b^2 equal to 1 and 9 a is a^2 is a integer(144,16 respectively) so a = 12,-12;4,-4 and b = 1,-1;3,-3. With these values of a and b we can make 8 ordered pairs.

a^2(b^2+1)+b^2(a^2+16)=448 => 2(ab)^2+a^2+16b^2=448 => 2l a^2 , Let be a=2c => 2(cb)^2+c^2+4b^2 = 112 => 2l c^2 => c=2d => 2(db)^2 +d^2 + b^2 = 28 => there are 2 solutions in positive (1,3) and (3,1) when a=12d (a,b)= (4,3) , (12,1) , (-4,3) (4, -3), (-4,-3), (-12,1), (12,-1), (-12,-1).

if (a,b) then (-a,-b),(a,-b),(-a,b) are solutions too,so we can assume a,b>=0

a^2(b^2+1)+b^2(a^2+16)=448

a^2(2b^2+1)=16(28-b²)

so because 2b^2+1 is odd

a^2 must br divided by 16,i.e. a must be divided by 4

putting a=0 b=2√7 isn't integer

a^2=16 : 2b²+1=28-b², i.e b^2=9, .i.e. , b=3

so there are 4 solutions (4,3) ,(4,-3),(-4,3),(-4,-3)

2a^2b^2<448 put b=1 ,so a^2<224,. i.e a<=12,so just to check a=8 and a=12

a=8 : 4(2b^2+1)=28-b^2, i.e. 9b^2=24, i.e .b=2√2/√3 or b=-2√2/√3 so this isn't integer

a=12: 9(2b^2+1)=28-b^2, i.e. 19b^2=19, i.e b=1 or b=-1

so there are 4 more solutions (12,1) ,(12,-1),(-12,1),(-12,-1)

So there are 8 solutions (4,3) ,(4,-3),(-4,3),(-4,-3),(12,1) ,(12,-1),(-12,1),(-12,-1)

A Diophantine equation like this one can be solved with relative ease if it is in factored form. So to begin, we will try to rewrite the equation in this way.

• Expand: $a^2b^2 + a^2 + a^2b^2 + 16b^2 = 448$

• Combine like terms and rearrange the LHS: $2a^2b^2 + 16b^2 + a^2 = 448$

• Now, Simon's Favorite Factoring Trick can be used: $2b^2(a^2 + 8) + (a^2 + 8) = 448 + 8$

• Simplify: $(2b^2 + 1)(a^2 + 8) = 456$ which is the factored form we want.

(If you don't know/understand Simon's Favorite Factoring Trick, do a quick web search for it. It is extremely useful in problems like this one.)

Define variables $m = 2b^2 + 1$ and $n = a^2 + 8$ , where $mn = 456$ , and $(m,n) \in \mathbb{Z}^+$ . Because $2b^2$ is even, $m$ must be a positive odd integer. Taking the prime factorization of 456, we find only four $m,n$ pairs where $m$ is odd: $(1,456), (3,152), (19,24),$ and $(57,8)$ . Now, we can use these pairs to find their analogous $a,b$ pairs which solve the equation.

$(m,n) = (1,456): a = \pm \sqrt 448$ . There is no such integer $a$ . Hence, there are $0$ possible pairs.

$(m,n) = (3,152): b = \pm 1$ and $a = \pm 12$ . Hence there are $4$ possible pairs.

$(m,n) = (19,24): b = \pm 3$ and $a = \pm 4$ . Hence there are $4$ possible pairs.

$(m,n) = (57,8): b = \pm \sqrt 28$ . There is no such integer $b$ . Hence, there are $0$ possible pairs.

Adding our results, we calculate a final answer of $0 + 4 + 4 + 0 = \boxed{8}.$

max value of a is 21(when b=0)

max value of b is 5 (when a=0)

checking for all values of b

(12,1) & (4,3) satisfy the above eq.

so 8 ordered pairs are possible.(4 for each)

Notation: LHS is the Left Hand Side of the equation, and RHS is the Right Hand Side of the equation.

We can rewrite the equation as $b^{2}(2a^{2} + 16) = 448 - a^{2}$ The term $448 - a^{2}$ must be greater than or equal its factors, so:

$(448 - a^{2}) >= (2a^{2} + 16)$ , thus $3a^{2} <= 432$ and then $a^{2} <= 144$ , so $|a| <= 12$ .

Keep in mind that LHS of the new equation is always even (since the term $(2a^{2} + 16)$ is always divisible by 2), and so RHS must be as well. If $|a|$ were to be odd, then $a^{2}$ would also be odd, and $448 - a^{2}$ would be odd. Thus, $|a|$ must be even.

$a = 0$ yields $b^{2}*16 = 448$ , or $b^{2} = 28$ . Has no solution over the integers, since 28 is not a perfect square.

$|a| = 2$ yields $b^{2}*24 = 444$ , or $2b^{2} = 37$ . LHS is even, RHS is odd, so this equation has no solution over the integers.

$|a| = 4$ yields $b^{2}*48 = 432$ , or $b^{2} = 9$ . Thus, $|b| = 3$ . We have 4 solutions from this.

$|a| = 6$ yields $b^{2}*88 = 412$ , or $b^{2}*22 = 103$ . LHS is even, RHS is odd, so this equation has no solution over the integers.

$|a| = 8$ yields $b^{2}*144 = 384$ . The LHS is a perfect square $((12b)^{2})$ , while 384 is not (since $19^{2} = 361$ and $20^{2} = 400$ ). So this equation has no solution over the integers.

$|a| = 10$ yields $b^{2}*216 = 348$ , or $b^{2}*18 = 29$ . LHS is even, RHS is odd, so this equation has no solution over the integers

$|a| = 12$ yields $304b^{2} = 304$ , so $b^{2} = 1$ , and thus $|b| = 1$ . We have another set of four solutions coming from this pair.

Our solutions are, then: $(-12, -1), (-12, 1), (12, -1), (12, 1), (-4, -3), (-4, 3), (4, -3)$ and $(4, 3)$ , giving us a total of 8.