2021/009) Solve for integers $(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c}) = 2$

Because of symmetry we have if $a,b,c$ is a solution then any permutation is a solution

Without loss of generality we choose $a \le b \le c$

So $(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c}) \le (1+\frac{1}{a})(1+\frac{1}{a})(1+\frac{1}{a})$

Or  $(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c}) \le (1+\frac{1}{a})^3$

So from given condition $(1+\frac{1}{a})^3 \ge 2$ or $(\frac{a+1}{a})^2 \ge 2$

The LHS is decreasing function of a and we have to see ratio of 2successsive cubes >2 as equal to 2 is not possibe

we have $\frac{64}{27} > 2$ and  $\frac{125}{64} <2$ giving $a < 4$

So we can have upto 3 values(as for some value we may not have a solution) for a . that is 1,2,3

putting a = 1 we get  $2(1+\frac{1}{b})(1+\frac{1}{c}) = 2$

or  $(1+\frac{1}{b})(1+\frac{1}{c})= 1$

as for b and c $\ge 1$ LHS is less than 1 this does not have a solution

putting a = 2 we get  $\frac{3}{2} (1+\frac{1}{b})(1+\frac{1}{c}) = 2$

or  $(1+\frac{1}{b})(1+\frac{1}{c}) = \frac{4}{3}$

or $3(b+1)(c+1) = 4bc$

or $3bc + 3b + 3c + 3 = 4bc$

or $bc -3b -3c -3 = 0$

or $b(c-3) -3 (c-3) = 12 $

or $(b-3)(c-3) = 12 = 1 * 12 = 2 * 6 = 3 * 4$

by taking one set a a time(as we have $b \le c$  we get (b,c) = (4,15) one solution (5,9) another solution and (6,7) 3rd solution

putting a = 2 we get  $\frac{4}{3} (1+\frac{1}{b})(1+\frac{1}{c}) = 2$

or  $4 (1+\frac{1}{b})(1+\frac{1}{c}) = 2 * 3$

or $2 (1+\frac{1}{b})(1+\frac{1}{c}) = 3$

or $2(b+1)(c+1) = 3bc$

or $2bc + 2b + 2c + 2 = 3bc$

or $(bc-2b -2c = 2$

or $(b-2)(c-2) = 6 = 1 * 6 = 2 * 3$

giving 2 sets of solutions (b,c) as (3,8) or (4,5)

So we have full solution in (a,b,c) as (2,4,15), (2,5,9),(2,6,7), (3,3,8), (3,4,5) or any permutation of any of the solutions