MHB Which Prime Numbers Make p^2 + 1007 Have Less Than 7 Divisors?

[lfdahl]

Determine all prime numbers $p$ such that the total number of positive
divisors of $A = p^2 + 1007$ (including $1$ and $A$) is less than $7$.

 

[kaliprasad]

Determine all prime numbers $p$ such that the total number of positive
divisors of $A = p^2 + 1007$ (including $1$ and $A$) is less than $7$.

Spoiler

P cannot be odd

reason if p is odd $p^2 $is 1 mod 8 so $p^2 +1007$ is divisible by $2^3$ so at least 4 *2 or 8 factors

so only candidate to be tested left is p =2

p =2 gives 1011 = 3 * 337 so 4 factor 1,3,337,1011

so only solution p = 2

 

[lfdahl]

Spoiler

P cannot be odd

reason if p is odd $p^2 $is 1 mod 8 so $p^2 +1007$ is divisible by $2^3$ so at least 4 *2 or 8 factors

so only candidate to be tested left is p =2

p =2 gives 1011 = 3 * 337 so 4 factor 1,3,337,1011

so only solution p = 2

Thankyou, kaliprasad for your participation. Clever answer!