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Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Number Theory of Primes.

There is a prime number p such that 16p+1 is the cube of a positive integer. Find p.

- is 307
- is 250
- is 840
- cannot be determined from the given information

**S**eries

Theory of Equations

Number Theory

But try the problem first...

Answer: is 307.

Source

Suggested Reading

AIME, 2015

Elementary Number Theory by Sierpinsky

First hint

Notice that 16p+1must be in the form \((a+1)^{3}=a^{3}+3a^{2}+3a\), or \(16p=a(a^{2}+3a+3)\). Since p must be prime, we either have p=a or a=16

Second Hint

p not equal to a then we have a=16,

Final Step

p\(=16^{2}+3(16)+3=307

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

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