Subtraction and Division Look Similar

First, note that $4-2 = 4\div 2 = 2,$ so the answer is $4+2=\boxed{6}.$ Let's see how to get that answer more methodically.

We multiply B for both sides, which gives $ab-b^2 = a$ . This will be tricky to factorize, so I will add B for both sides of this equation. This gives $ab-b^2+b-a = b$ . Factorize the latter form gives $a(b-1)-b(b-1) = b = (a-b)(b-1)$ Now wrote $a$ in terms of $b$ , which gives $a = \frac{b}{b-1}+b = b+(1+\frac{1}{b-1})$ When you see the form $a = b+1+\frac{1}{b-1}$ , no other whole number will make the fraction being whole number (as A must also be whole number) except -1 and 1 (division by zero is not allowed). Here, we will get $b-1 = \pm1$ or $b = 0, 2$ . Again, as the division by zero is not allowed, $b \neq 0$ , so $b = 2$ is the only solution, and when solving the equation, gives only $(a, b) = (4,2)$ as the only answer, and $a+b = 6$

We know that both A and B are integers. Then A-B is an integer, and therefore A/B is an integer. Since A/B is an integer, we know that B divides A. Then there exists an integer k such that Bk=A. Substituting Bk for A in our equation, we get Bk-B=Bk/B. Then Bk-B=k. We can factor this to obtain B(k-1)=k. Now note that k-1 is not 0. If it were, we would obtain the contradiction that B(0)=1. Then dividing by k-1, we get B=k/(k-1) and subsequently B=1+1/(k-1). Now, since both B and 1 are integers, we know that 1/(k-1) is an integer, which means that k-1 divides 1. Since the only divisors of 1 are 1 and -1, we know that either k-1=1 or k-1=-1. We can reject the latter since it will result in k=0 and B=0 (since B(k-1)=k would mean B(-1)=0), which is not possible since B divides A. Then k-1=1 and k=2. Then B(k-1)=k tells us that B=2, and since Bk=A, we know that A=Bk=2*2=4. Therefore A+B=4+2=6.

[Solution using pythagorean triplets] $a - b = \frac{a}{b}, \text{ where } a, b \in \mathbb{Z^+} \\ \Rightarrow ab - b^2 = a \\ \Rightarrow b^2 - ab + a = 0$ Take the discriminant of the quadratic: $Δ = (-a)^2 -4\cdot a \\ \Rightarrow Δ= a^2-4a \\ \stackrel{(*)}{\Rightarrow} r^2 = a^2 - 4a , \text{for some } r \in \mathbb{Z^+} \\ \Rightarrow r^2 = a^2 - 2\cdot2\cdot a +4 -4 \\ \Rightarrow r^2 = (a-2)^2 -4 \\ \Rightarrow (a-2)^2 = r^2 + 2^2$ Which means that $\Big((a-2), r, 2\Big)$ is a pythagorean triplet and for every pythagorean triplet $A, B, C$ (where $A^2 = B^2 + C^2$ ) it true that $(A, B, C) = \Big(k(m^2 + n^2), k(m^2 - n^2), k\cdot 2mn\Big)$ for some positive integers $k, m, n$ . Now we have to consider two possible scenarios:

First scenario: $2 = k\cdot 2mn$ : $\Rightarrow 1 = kmn \stackrel{k,m,n\in\mathbb{Z^+}}{\Longrightarrow} (k = 1 \text{ and } m=1 \text{ and } n=1) \\ \Rightarrow \Big((a-2), r, 2\Big) = (2, 0, 2) \\ \Rightarrow (a, b) = \Bigg((2+2), \Big(\frac{a \pm \sqrt{Δ}}{2}\Big)\Bigg) \\ \stackrel{Δ=0}{\Longrightarrow}\boxed{(a, b) = (4, 2)}$

Second scenario: $2 = k(m^2-n^2)$ : $\Rightarrow 2 = k(m+n)(m-n) \\ \stackrel{(m+n)\geq(m-n)}{\Longrightarrow} \Big[k=1 \text{ and } (m+n)=2 \text{ and } (m-n)=1\Big] \text{ Or } \Big[k=2 \text{ and } (m+n)=1 \text{ and } (m-n)=1\Big] \\ \Rightarrow \Big[k=1, m=1.5, n=0.5\Big] \text{ Or } \Big[k=2, m=1, n=0\Big]$ But in the first case where $(k, m, n) = (2, 1.5, 0.5) \notin \mathbb{Z^+}$ so we get no solution. In the second case where $(k, m, n) = (2, 1, 0)$ we get: $\Big((a-2), r, 2\Big) = (2, 0, 2) \\ \Rightarrow (a, b) = \Bigg((2+2), \Big(\frac{a \pm \sqrt{Δ}}{2}\Big)\Bigg) \\ \stackrel{Δ=0}{\Longrightarrow}\boxed{(a, b) = (4, 2)}$

$(*)$ : We expect the discriminant to be a perfect square of an integer so that $b$ will have integer values.

$b\neq 0, \quad a-b=a\div b \quad \Rightarrow ab-b^2=a \quad \Rightarrow ab-b^2-a+1=1;$

$\Rightarrow a(b-1)-(b+1)(b-1)=1 \quad \Rightarrow (b-1)(a-b-1)=1;$

$a,b \in \mathbb{Z} \Rightarrow \{ b-1=1 \wedge a-b-1=1 \} \vee \{ b-1=-1 \wedge a-b-1=-1 \}$

$\text{Which Gives us}, \boxed{b=2} \wedge \boxed{a=4},$

$\scriptsize \text{The second condition gives us}, b=0, \text{which is inappropriate!!!}$

$\Box$

4 - 2 = 4/2. therefore 4+2 = 6, A=4 & B = 2

4-2=2 4➗2=2 (2+4=(6)).

4-2= $\frac{4}{2}$

A - B = A / B

A + B = (A / B) + B^{2}

A + B = A - B + B^{2}

B = - B + B^{2}

B^{2} - 2B = 0

B (B - 2) = 0

B = 0 V B = 2

A/B so B is not equal to 0, so B = 2

Then,

A - 2 = A / 2

2A - 4 = A

A = 4

So, A = 4 and B = 2

A + B = 4 + 2 = 6.

we see that b cannot be equal to 1 and also zero . further b-1 should divide b^2 which is only possible when b-1 divides -1 . as b cannot be zero it has to be two and a 4 which leads to their sum equal to 6

First, $b \not = 0$ . Next we have that $a-b \in \mathrm{Z}$ , so we have $a = kb$ for some integer k. We rewrite $kb-b = k$ , so $b(k-1)=k$ , and $b = k/(k-1)$ . The two integer solutions to this equation are clearly $k=0$ and $k=2$ , leading to $b=0$ and $b=2$ , and thus $a=0$ and $a=4$ . The only solution that satisfies $b \not = 0$ is $\{2, 4\}$ giving a sum of $6$ .

A - B = A/B..... a whole number.
Dividing both sides by B,
(A/B) - 1 = A/B^2......a whole number. Rearranging gives,
(A/B^2)(B - 1) = 1.
This is possible only if,
(A/B^2) = 1, and B - 1 = 1,
so B = 2 and A = B^2 = 4.
A+B = 6.

6 is a perfect number too, that's creepy.

a-b = a/b => as a, b are integers therefore b | a => a = bk, k = some +ve integer. bk-b = k => b(k-1) = k as gcd(k-1, k) = 1, b|k and b is an integer => k can only be 2. so b = 2, a = bk = 4. => a+b = 6

How I did this? Guess and check.

$A-B=\dfrac{A}{B} \implies A(B-1)=B^2 \implies (B-1) \mid B^2 \implies B=2$ .

So, $A=4$ .

We have, $\boxed{A+B=6}$ .

And why there is only one solution? Because, $(B-1) \mid B^2 \implies B=2$