$\begin{aligned} a + b &= ab \\ a &= ab - b \\ a &= b(a-1) \\ \frac{a}{a-1} &= b \end{aligned}$

(There's no possible division by zero here; $a \neq 1$ since that would imply $1 + b = b .)$

Since $a$ is a positive integer, $b$ is from the set $\{\frac{2}{1}, \frac{3}{2}, \frac{4}{3}, \frac{5}{4}, \frac{6}{5}, \ldots\}.$

However, with the exception of $2 \div 1,$ consecutive positive integers when divided have a remainder of 1, so $b$ cannot be an integer except in the case of $\frac{2}{1} .$ This implies the only positive integer solution is $a = b = 2 .$

Add $1-a-b$ to both sides of the equation to get $ab - a - b + 1 = 1 \\ (a-1)(b-1) = 1$

Note that $a-1, b-1$ are integers, and the only way for the product of two integers to equal $1$ is for $a-1 = -1 = b-1 \qquad \text{ or } \qquad a-1 = 1 = b-1$ Either way, we have $a=b$ , which is disallowed in the problem, so the answer is $\boxed{\text{no}}$ .

Dividing both sides by $a\times b$ gives $\dfrac 1a + \dfrac 1b = 1$ .

Because $a$ and $b$ are distinct, without the loss of generality, we can assume that $a<b$ , then $\frac1a > \frac1b \quad \Rightarrow \quad \frac1a + \frac1a > \frac 1a + \frac1b = 1 \quad \Rightarrow\quad a < 2.$

This means that $a$ must be equal to 1 only, then $\dfrac1a + \dfrac1b = 1\Rightarrow \dfrac1b = 0$ which is absurd because the ratio of 2 positive integers cannot be equal to 0. Hence, there's no solution.

Without loss of generality, suppose $b > a$ . Consider two possible cases:

Case I : $a = 1$ . Then $\begin{aligned} a \times b & = 1 \times b \\ & = b \\ & < 1 + b = a + b. \end{aligned}$

Case II : $a \geq 2$ . Then $\begin{aligned} a \times b & \geq 2 \times b \\ & = b + b > a + b. \end{aligned}$

In either case, $a \times b \not = a + b$ .

well, the thing is I have no fancy equation to state I simply want to explain how my number instincts reacted so if that's what you're looking for you might be disappointed. ok with that being said here's how I solved this one. at first, I thought "ok, let's try to find a number that works!!" but I realized it's impossible the higher the number goes the further the product from the sum!! and no decimals would work so I selected no and got it right!!(~˘▾˘)~

Solving $a + b = ab$ for $b$ gives $b = \frac{a}{a - 1}$ which is $b = \frac{a - 1 + 1}{a - 1}$ or $b = 1 + \frac{1}{a - 1}$ .

In order for $b$ to be an integer, $a - 1$ must divide evenly into $1$ , and since $a$ is also an integer and since the only integer factors of $1$ are $1$ and $-1$ , $a - 1 = 1$ or $a - 1 = -1$ , which solves to $a = 2$ or $a = 0$ . Since $a$ must also be positive, $a = 2$ .

If $a = 2$ , then $b = 1 + \frac{1}{2 - 1} = 2$ , which means if $a$ and $b$ are integer solutions then they are not distinct.

Suppose b>a, divides b on both sides

We have $\frac{a}{b}$ +1 = a

or $\frac{a}{b}$ = a-1

The left hand side is a proper fraction but the right hand side is an integer .

So NO .

Without any loss of generality, let's assume that $a\geq b$ , where $b>2$ . Hence, we have $a+b=ab>2a,$ which implies $b>a,$ yielding a contradiction.

We can make the problem quite a bit easier by checking if the equation balances in certain circumstances.

For example, if we let one of $a$ or $b$ be even and the other odd, then looking at the equation we see that this gives us $\text{even} + \text{odd} = \text{even}\times\text{odd}$ which is clearly false. Similarly, setting both $a$ and $b$ as odd we get that $\text{odd} + \text{odd} = \text{odd}\times\text{odd}$ which is also impossible. This means that both $a$ and $b$ must both be even. We will use this fact later.

Now rearranging the equation:

$a + b = a \times b \rightarrow b = \dfrac{a}{a - 1}$

If we let $a = 2m$ and $b = 2s$ then we get that $2s = \dfrac{2m}{2m - 1} \rightarrow s = \dfrac{m}{2m - 1}$ . Note that for integers $m > 1$ , $2m - 1 > m \implies 0 < s < 1$ and therefore $b$ will not be an integer. This is of course unless $m = 1$ in which case we get the solution $2 + 2 = 2\times 2$ which we can ignore.

Here is a graph from which it can be seen the only two lattice points are (2,2) and (0,0). The dotted lines are asymptotes where the graph is approaching a=1 in one direction and b=1 in the other, but this never happens.

It's very easy!

Since a+b=ab

ab-a-b=0

ab-a-b+1=1

a(b-1)-(b-1)=1

(a-1)(b-1)=1=1 * 1=(-1) * (-1)

Therefore, a and b be 0;0 or 2;2

Passing the 'a' to the other side and putting it on evidence we can see that (b-1) needs to be a divisor of b. The only integer that satisfies that is '2'. So b (and therefore a) are both equal to 2.

Without loss of generality, $a < b$ .

Case 1: $a = 1$ . Then $b = a*b < a + b$ .

Case 2: $a \geq 2$ . Then $a + b < 2*b \leq a * b$ .

• We can prove that there is just one case of two possible integers equal to each other being equal to 2 referring to case that satisfy the condition 2+2=2x2 by writing a series of identities that are true and may induct the case; otherwise in the final step there is no other solution so the only scenario stays true to the integer: 2. a + b = a x b (a + b)(a - b) = a x b (a- b) a^2 - b^2 = a^2b - ab^2 a^2 - a^2b = b^2 - ab^2 a^2(1 - b) = b^2(1 - a) We have assumed that a>b <=> a-b>0, and also this means a^2/b^2>0 and b =/= 0, or 1 --since for b=0 we have: a + 0 = 0 This compromises the condition to get a=/=0; we could also had initially b stating it is a positive integer and not proved it for a. At the same time if b = 1, b-1=0 and a^2(1-b)=b^2(1-a): 0 = 1 - a <=> a= -1; again not a >0 as in condition of problem. Therefore we can state the last identity: a^2/b^2 = (1-a)/(1-b) From this we have a=b to get the ratios equal in both sides otherwise we would have a ratio =/= 1 : impossible..., because we stated earlier that a^2/b^2>0 <=> (1-a)/(1-b)>0 so this entails only the case where a=b and both greater than 0 otherwise if a=/=b (1-a)/(1-b) =/=1 and <0, but we started in the right of the identity firstly with assuming that a^2/b^2>0 which derives directly from the conditions of the problem a and b are positive integers so both of their squares are greater than zero. Now we can find if there is any solution to this problem by solving and substituting in the identity a^2(1-b)=b^2(1-a) the latest of this: (1-a)/(1-b) = 1 <=> 1 -a = 1/(1-b) so a=b and a or b =/= 1 or 0 and are positive integers, so 1-b= 1/(1-b) and we get b(b-2)=0 <=> b=0 or b=2 This means the only possible case is letting a=b=2(Q.E.D.)

Take modulo $a$ of both sides:

$a+b \equiv ab \mod a$

$0+b \equiv 0\cdot b \mod a$

$b \equiv 0 \mod a$

This means $b=ka$ for some integer $k$ .

Taking modulo $b$ of both sides similarly yields $a=lb$ for some integer $l$ .

Substituting for $a$ in $b=ka$ :

$b=k(lb) \Rightarrow 1=kl \Rightarrow l=1/k$

$l$ is an integer, so $1/k$ is an integer. However, $k$ is also an integer

$\therefore k=1 \Rightarrow b=1\cdot a = a$

So $a$ and $b$ must be equal.

a+b = ab; So a = (b/b-1) implies that a can take integer values only when b = 2 , for all other values of b greater than 2, a will be decimal which is not allowed as per the question.

• a+b=a*b
• a+b/a*b=1
• 1/b+1/a=1
• we will not get 1/a+1/b =1 for any number except 2. so there is no possible solution for positive integer...

a+b=a*b can be treated as x+y-xy=0 so simply we have to find its intersection with x=y line which is at (0,0) and (2,2) hence they are not distinct positive integers

if
$a+b = ab$
then
$a = b(a-1)$
and
$\frac{a}{a-1} = b$

A little more rearranging gives
$\frac{1}{a-1} = b-1$
From here it is easy to see that the $\frac{1}{a-1}$ part will never be an integer after 2 and so there are no two distinct integer solutions

This question can be solved by considering some sample values:

i. The solution can not have one of the values as 1, since addition increments the other value but multiplication doesn't.

ii. Any other value apart from (2, 2) will definitely show massive difference between the sum and product (e.g., the lowest set 2+3=5 and 2×3=6), and this difference widens with higher set of values.

We can rearrange $a+b=a\times b$ to $\frac{a}{a-1}=b$ .

for $b$ to be an integer, $a-1$ must be a factor of $a$ . The only number for which this is the case is $2$ (as all other cases would remainder $1$ ).

It's obvious that solutions $(a,b) = (1,1)$ , $(a,b) = (1,b)$ or $(a,b) = (a,1)$ are not possible because of additive and multiplicative property of 1. Now, let's assume there are positive integers $a$ and $b$ not less than $2$ such that the expression holds. Let it be:

$\begin{aligned} a &= 2 + x \\ b &= 2 + y, \end{aligned}$

where $x$ and $y$ are some non-negative integers. Then, the following must also hold:

$\begin{aligned}a + b &= ab \\ (2 + 2) + x + y &= (2 \cdot 2) + 2(x+y) + xy \\ -(x+y) &= xy \end{aligned}$

But since $x$ and $y$ are non-negative integers, left-hand side of the expression must be non-positive and right-hand side non-negative. The expression, thus, only holds when $(x,y) = (0,0)$ implying that $(a,b) = (2,2)$ is the only solution.

If $a$ and $b$ are distinct & positive this means that they are not the same and must be larger than zero. Then you can assume that if $a$ > $b$ (the reverse could be argued) then $a = b + n$ , where $n$ must also be positive and larger than zero (to make $a$ and $b$ distinct) .

Substitute this to the equation above and you obtain a quadratic equation in terms of $b$ and therefore a solution in terms of $n$ : $b = \frac{2-n\pm\sqrt{n^2+4}}{2}$ .

Inspect the determinant of the quadratic, in order for $b$ to be an integer at the square root of this determinant must be rational (then of course the whole numerator must be a multiple of 2 but need not go that far). For the determinant to be rational $n^2 + 4$ must therefore equal a square number i.e. 4, 9, 16, 25, etc.

Since n must be a positive integer which is larger than zero the minimum square number it can equal is 9. From 9 and beyond the difference between square numbers is larger than 4 thus making $n$ irrational and so the determinant irrational.

Suppose $a+b = ab$ and divide both sides by $a$ . $1 + \frac{b}{a} = b$ . $\frac{b}{a}$ must be integer, therefore $a$ divides $b$ (write this as $a | b$ ). By symmetry, we also have $b | a$ . But integers under the division relation form a partially ordered set (a poset) so if $a|b$ and $b|a$ , then $a=b$ for all solutions.

I don't need any equation to prove myself. I'm confident of my answer. There doesn't exist any other integer except 2.

Did something change in the app? It happened again.. I answer a question correctly to continue my streak and it does not register correctly.