Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Cube of Positive Integer.

## Cube of Positive Numbers – AIME I, 2015

There is a prime number p such that 12p+1 is the cube of positive integer.Find p..

- is 107
- is 183
- is 840
- cannot be determined from the given information

**Key Concepts**

Algebra

Theory of Equations

Number Theory

## Check the Answer

Answer: is 183.

AIME, 2015, Question 3

Elementary Number Theory by David Burton

## Try with Hints

\(a^{3}=12p+1\) implies that \(a^{3}-1=12p\) that is (a-1)(\(a^{2}\)+a+1)=12p

a is odd, a-1 even, \(a^{2} +a+1 odd implies a-1 multiple of 12 that is here =12 then a=12+1 =13

\(a^{2}+a+1=p implies p= 169+13+1=183.

## Other useful links

- https://www.cheenta.com/cubes-and-rectangles-math-olympiad-hanoi-2018/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s