Question

Math :
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Given that p and 8p²+1. Find p.

Answer 1

Question :

Given that p and 8p² + 1. Find p.

Solution :

Any perfect square number, can be expressed in the form 3k or 3k + 1 where k is a constant term.

EXAMPLES :

• 1² → 1 → 3(0) + 1
• 2² → 4 → 3(1) + 1
• 3² → 9 → 3(3)
• 4² → 16 → 3(5) + 1
• 5² → 25 → 3(8) + 1
• 6² → 36 → 3(12)

From the pattern, we observe that any perfect square number, a² is in the form 3k, when it is divisible by 3 and is in the form 3k + 1, if it is not divisible by 3.

• Suppose in 8p² + 1, p is not divisible by 3, then p² is in the form 3k + 1.

Let us substitute p² for 3k + 1.

⇒ 8p² + 1

⇒ 8(3k + 1) + 1

⇒ 24k + 8 + 1

⇒ 24k + 9

• For any integer value k, 24k + 9 will always be divisible by 3, since 24k and 9 are divisible by 3.

If p² is not divisible by 3, then 8p² + 1 is always divisible by 3. But from the question, we know that 8p² + 1 is a prime number and the only prime number divisible by 3, is 3 itself. p will not take an integer value for 8p² + 1 to be divisible by 3.

• Now that we know that p is divisible by 3, p² will take the value of 3k

Substituting p² for 3k,

⇒ 8p² + 1

⇒ 8(3k) + 1

⇒ 24k + 1

∴ For any integer value for k, 24k + 1 will always not be divisible by 3.

• From the above, we can conclude that p² is divisible by 3, hence making p also be divisible by 3.
• Given in the question, p is prime. The only prime number which is divisible by 3 is 3 itself.

Therefore, p = 3.

Verification :

Substituting p² for (3)² in 8p² + 1,

⇒ 8(3)² + 1

⇒ 8(9) + 1

⇒ 72 + 1

⇒ 73

73 and 3 are both prime numbers. Hence p = 3.