Largest consecutive integers with no prime factors except $2$, $3$ or $5$?

The number $180$ has a special property. Its prime factors are only $2$, $3$, and $5$. However the number $220$ does not have this special property because one of its prime factors is $11$.

In the first one-hundred natural numbers, the numbers with this property are plentiful. They cover almost half of that range. However, into the thousands, or tens of thousands, they get increasingly rare, and you might go hundreds of integers without finding one. As far as I know, $80$ and $81$ are the two largest consecutive numbers that both have this property — $80 = 2^4 \times 5$ and $81 = 3^4$. I have tested numbers up to one million, and have not found another pair of numbers like this.

But, is it possible to mathematically prove this? Has anyone ever tried to do it before? All evidence points to the fact that $80$ and $81$ are the largest pair of integers with this property, but has it ever been proven before?

Solution 1:

Only one of the two consecutive numbers can be divisible by $2$. Similarly, only one is divisible by $3$ and only one is divisible by $5$. Depending on how we pair the prime powers we therefore have six possibilities.

The cases $p^Aq^B=r^C-1$ can all be solved in the same way.

$2^A3^B=5^C-1 $

If $A\ge4$ then modulo $16$ we have $C\equiv 0\pmod 4$ and then $5^C-1$ has a factor of $13$.

If $B\ge2$ then modulo $9$ we have $C\equiv 0\pmod 6$ and then $5^C-1$ has a factor of $31$.

The remaining values of $A,B$ give the solutions $\{4,5\},\{24,25\}$.

$2^A5^B=3^C-1 $

If $A\ge5$ then modulo $32$ we have $C\equiv 0\pmod 8$ and then $3^C-1$ has a factor of $41$.

If $B\ge2$ then modulo $25$ we have $C\equiv 0\pmod {20}$ and then $3^C-1$ has a factor of $11$.

The remaining values of $A,B$ give the solutions $\{2,3\},\{8,9\},\{80,81\}$.

$3^A5^B=2^C-1 $

If $A\ge2$ then modulo $9$ we have $C\equiv 0\pmod 6$ and then $2^C-1$ has a factor of $7$.

If $B\ge2$ then modulo $25$ we have $C\equiv 0\pmod {20}$ and then $2^C-1$ has a factor of $31$.

The remaining values of $A,B$ give the solutions $\{1,2\},\{3,4\},\{15,16\}$.

The cases $p^Aq^B=r^C+1$.

$2^A\times3^B=5^C+ 1$

Modulo $4$ we have $2^A\times3^B\equiv 2\pmod 4$. Therefore $A=1$ and $2\times3^B=5^C+ 1.$

If $B\ge2$ then modulo $9$ we have $C\equiv 3\pmod {6}$ and then $5^C+1$ has a factor of $7$.

The remaining values of $B$ give the solution $\{5,6\}$.

$2^A\times 5^B=3^C+ 1$

Modulo $4$ we have $2^A\times3^B\equiv 2\pmod 4$. Therefore $A=1$ and $2\times5^B=3^C+ 1.$

If $A\ge3$ then modulo $8$ we have $3^C\equiv -1\pmod {8}$ which is impossible.

If $B\ge2$ then modulo $25$ we have $C\equiv 10\pmod {20}$ and then $3^C+1$ has a factor of $1181$.

The remaining values of $A,B$ give the new solution $\{9,10\}$.

$3^A\times 5^B=2^C+ 1$

The cases $AB=0$ have been covered in previous equations. So we can suppose $AB>0$ when the equation is impossible modulo $15$.