Determine-all-the-perfect-squares-on-form-p-n-1-where-p-is-a-prime-number-and-n-a-positive-integer-

Question Number 154012 by mathdanisur last updated on 13/Sep/21

Determine all the perfect squares on

form p^{n} + 1 where p is a prime number

and n a positive integer .

Commented by Rasheed.Sindhi last updated on 13/Sep/21

For some cases

p^{n} + 1 = q^{2} \Rightarrow q^{2} - 1 = p^{n}

(q - 1) (q + 1) = p^{n}

n = 1 : (q - 1) (q + 1) = p

q - 1 = 1 \land q + 1 = p

q = 2 \land p = 3

q^{2} = 4

n = 2 : (q - 1) (q + 1) = p^{2}

case 1 : q - 1 = 1 \land q + 1 = p^{2}

q = 2 \land p^{2} = 3 (false)

case 2 : q - 1 = q + 1 = p (false)

For n = 2 no p

n = 3 : q^{2} - 1 = p^{3}

p^{3} + 1 = q^{2}

(p + 1) (p^{2} - p + 1) = q^{2}

p + 1 = p^{2} - p + 1

p^{2} - 2 p = 0

p (p - 2) = 0

p = 0 (false) \lor p = 2 (true)

p = 2 \Rightarrow q^{2} = p^{3} + 1 = 2^{3} + 1 = 9

q^{2} = 9

Commented by mathdanisur last updated on 13/Sep/21

Very nice Ser, thank you

Answered by Rasheed.Sindhi last updated on 14/Sep/21

p^{n} + 1 = q^{2}

p^{n} = (q - 1) (q + 1)

q - 1 = p^{m} \Rightarrow q + 1 = p^{n - m}

p^{n - m} - p^{m} = (q + 1) - (q - 1) = 2

p^{n - m} - p^{m} = 2

(T w o p o w e r s o f a p r i m e h a v i n g

d i f f e r e n c e 2)

2^{2} - 2^{1} = 2 \Rightarrow q - 1 = 2 \land q + 1 = 4

\Rightarrow q^{2} - 1 = 8 \Rightarrow q^{2} = 9

(O n l y t h e s e t w o p o w e r s o f 2 h a v e

{d i f f e r e n c e 2)}^{★}

3^{1} - 3^{0} = 2 \Rightarrow q - 1 = 3^{0} \land q + 1 = 3^{1}

\Rightarrow q^{2} - 1 = 3 \Rightarrow q^{2} = 4

(O n l y t h e s e t w o p o w e r s o f 3 h a v e

{d i f f e r e n c e 2)}^{★ ★}

p > 3 : p^{m} - p^{n} > 2 f o r a n y m, n

Commented by Rasheed.Sindhi last updated on 14/Sep/21

^{★} 2^{1} - 2^{0} = 1 (\times)

2^{2} - 2^{1} = 2 (✓)

2^{m} - 2^{n} > 2 (\times) f o r m, n > 2 \land m \neq n

^{★ ★} 3^{1} - 3^{0} = 2 (✓)

3^{m} - 3^{n} > 2 (\times) f o r m, n > 1 \land m \neq n

Commented by mathdanisur last updated on 14/Sep/21

very nice S er thanks