+0

Math Question

0
138
7

Prove that 4 is the only perfect square that is one more than a prime number

Jan 12, 2021

#1
+8
0

I would say the answer is, No. I would say that because if you check the next perfect square, 8, it is one above 7 which is a prime number. (correct me if I'm wrong)

Jan 13, 2021
#2
0

8 is not a perfect square.

Guest Jan 13, 2021
#3
+8
0

I am very sorry. I am a bit rusty on my elementary math... it has been about a year since I've had a refresher for it.  Let me fix that mistake and take a bit of time into that. I would say it is the only perfect square that is one above a prime number I went all the way to 263.

Arrakis  Jan 13, 2021
#4
0

It's not a yes or no answer it's telling to prove how, and I should add that it's supposed to use factoring.

Guest Jan 13, 2021
#5
+112827
+1

A perfect square is of the form x^2 where x is a positive whole number

One less than a perfect square is    x^2 -1

\(x^2-1=(x-1)(x+1)\)

If this is a prime number then the smaller factor must be 1

x-1=1

x=2

x^2-1 = 3

Hence 3 is the only prime number that is followed by a perfect square.

Jan 13, 2021
#6
+1

Can you explain how you got to x-1=1

Guest Jan 14, 2021
#7
+112827
+1

I shall try

A perfect square is of the form x^2 where x is a positive whole number

One less than a perfect square is    x^2 -1

\(x^2 -1\)     is the difference of 2 squares.

difference means subtract and the two squares are   x^2    and    1^2

You need to memorize this factorization.

\(\boxed{a^2-b^2=(a-b)(a+b)}\)

so        \(x^2-1^2=(x-1)(x+1)\)

If this is a prime number then the smaller factor must be 1

the factors are  (x-1) and (x+1)

since x is a positive whole number,  the smallest one will be x-1

therefore  x-1 must equal 1

so x must equal 2

Melody  Jan 15, 2021