Let $d=\gcd(a,b)$, $a=dx$, $b=dy$, with $\gcd(x,y)=1$. Then, since now $\operatorname{lcm}[a,b] = dxy$, the relation becomes \[1+9xy+9(x+y)=7dxy.\] This forces $d>1$ anyways. We need then have $xy \mid 9(x+y)+1$, so $9(x+y) + 1 = kxy$, for some natural $k\geq 1$. But then the relation becomes $(k+9)xy = 7dxy$, thus $7\mid k+9$, forcing $k\geq 5$. Therefore $9(x+y) + 1 \geq 5xy$, or $(5x-9)(5y-9)\leq 86$. We are left with a simple casework. The only special case is when, say, $x=1$; this leads to $9(1+y) + 1 = ky$, or $10 = (k-9)y$, with the only possibility $k=19$, $y=1$, thus $d=4$ and indeed $(x,y) = (4,4)$ is a solution. For the other cases we have $2\leq x,y\leq 19$, and there are very few possibilities to consider.

let d=gcd(a,b), a=dx and b=dy ( gcd (a,b) =1 ). lcm [a,b] =abd; the equation becomes : 1 + 9xy +9x +9y =7dxy Obvious that: 7dxy = 1+9xy+9x+9y <= xy+9xy+9xy+9xy = 28xy ................d<=4 (equality holds when x=y=1; in which case (a,b)= (4,4) ) So, d={2,3} By resolving the cases (the second one is very simple) we get the other solution (38,4)

Thats what I got by solving it

Good solutiont thank a lot

As obvious similar questions here $gcd(m,n)+lcm(m,n)=m+n$ then $m|n$ $n|m$

Here is my solution Let $d=gcd(a,b)$, $a=dx$ ,$b=dy$ Now $lcm[a,b]=dxy$ the relation becomes $$1+9xy+9(x+y)=7dxy.$$$$\implies \frac{1}{xy}+9+\frac{9}{y}+\frac{9}{x}=7d$$Now $x, y\geq 1$ and x and y are integers so $l. H. S\leq 28$.hence We get $d\leq 4$ $case 1:d=4$ The solution is obvious as $(x, y) =(1,1)$ $case 2:d=3$ If $x, y\geq 2$ we get $$\frac{1}{4}+9+\frac{9}{2}+\frac{9}{2}\leq 19$$So only case left is when $(x, y) =1,2$ But this gives value greater than 21.hence no solutions $case 3:d=2$ If $x\geq 3$,$y\geq5$then we get $$\frac{1}{15}+9+\frac{9}{3}+\frac{9}{5}< 14$$So we have to check cases when $x\leq 3$, $y\leq 5$ So easy bash gives solutions as $(x, y) =(2,19)$ $case 4:d=1$ Not possible as $\frac{1}{xy}+9+\frac{9}{y}+\frac{9}{x} >9$. So all solutions are $(a, b) =(4,4),(38,4)$ and it's permutations.

First, let's accept $a=d*x$ and $b=d*y$. Also, d is not equal to 1, that means a and b are not relatively prime. So, let's solve. $d+9dxy+9dx+9dy=7*d^2*x*y$ We can divide this equality into d. $1+9xy+9x+9y=7dxy$ (1) Right handside is divisible by $x*y$ so, lefthandside must also. $xy \mid 9xy.$ $xy \mid 9*(x+y)+1$. Let's say $9*(x+y)+1=kxy$ Use it in (1) equality. $kxy+9xy=7dxy$ so, $k+9=7d$ $k=7*d-9$ . d can be 1,2,3,4. Because, $7dxy<=xy+9xy*3=28xy$. And $d<=4$. Let's search the cases; i case) IF d=4; k=19 $19*x*y=9*x+9*y+1$ it is easy to see $x=y=1$ and for d=4 ======> $(a=4 b=4)$ ii case) IF d=3; k=12 $12*x*y= 1+9x+9y$ $(9x+1)*(9y+1) = -69xy$ we know that d is positive and x,y,a,b can not be negative numbers. NO solution. iii case)IF d=2; k=5 $5xy=1+9x+9y$ from the same way we can get a solution as x=2, y=19 =====>$(a=4 b=38)$ iv case)IF d=1; there will be no solution. Because there will be a situation as we met (negative numbers ) . SOLUTIONS ARE: (a,b)======> (4,4), (4,38),