Question

((m!)^2) + 23 is a perfect square, how many values can m take

Answer 1

Answer:

no such value of m

Step-by-step explanation:

((m!)²) + 23 is a perfect square

Let say m! = a

a² + 23  =  b²

=> b > a

Case 1

if b = a + 1

then

a² + 23 = (a + 1)²

=> a² + 23 = a² + 1 + 2a

=> a = 11  & b = 12

11 can not be m!

case 2

if b = a + 2

then

a² + 23 = (a + 2)²

=> a² + 23 = a² + 4 + 4a

=> a is not an integer

not possible

Case 3

if b = a + 3

then

a² + 23 = (a + 3)²

=> a² + 23 = a² + 9 + 6a

=> a is not an integer

not possible

case 4

if b = a + 4

then

a² + 23 = (a + 4)²

=> a² + 23 = a² + 16 + 8a

=> a is not an integer

not possible

case 4

if b = a + 5

then

a² + 23 = (a + 5)²

=> a² + 23 = a² + 25 + 10a

=> (a + 5)² > a² + 23

not possible

& no further Solution

So no such value of m Exists for which ((m!)²) + 23 is a perfect square

if it wold have been 2³ instead of 23

them m = 1

1! = 1

1² + 2³ = 9 = 3²