Interesting 3

Firstly, $p=n^2-1\implies p=(n+1)(n-1)$

Which only works for $n=2\implies p=3$

• One and two have more letters than their numerical values too.

• 12 is a multiple of 3 and one more than a prime (11).

• 11 is a prime one less than a multiple of 4.

Simply check if each condition is true:

3 is the only multiple of 3 that is one more than a prime.

Let's check 6.

• 6 is a multiple of 3.
• 6 is one more than 5.
• 5 is a prime number.

False.

3 is the only integer whose spelling has more letters than its numerical value.

Let's check 1.

• 1 is spelled "one".
• "One" has 3 letters.

False.

3 is the only prime that is one less than a multiple of 4.

Let's check 7.

• 7 is one less than 8.
• 8 is a multiple of 4.

False.

3 is the only prime that is one less than a perfect square.

This must be true because:

• 1 out of all 4 conditions must be true for this problem to be solvable.
• All other conditions are false.

True.

Answer: 3 is the only prime that is one less than a perfect square.

Rather than showing how the other three statements are false, I took the statement "3 is the only prime number that is one less than a perfect square. Algebraically, one less than a perfect square can be written as: x^2-1=0 If we factor the above expression, we have: (x+1)(x-1)=0 for the number in question to be prime, either (x+1) or (x-1) has to equal 1. If we omit the case where x=0 (since neither 1 nor -1 Is prime), the only other case is when x=2, yielding the solution set of (1,3). Therefore, this shows that 3 is the only prime number that is one less than a perfect square.