Find a counter example to the statement?

Question

"The square of a number added to the sum of the number and 5 is a prime number" work it out.

Answer 1

Let the number be x and the resultant number be y.

The equation is therefore,
x^2 + (x + 5) = y
y = x^2 + x + (0.5)^2 + 5 - (0.5)^2
y = (x + 0.5)^2 + 4.75
Minimum point of curve --> (-0.5, 4.75)

The statement is definitely untrue because for all real values of x, y may not be a prime number. y can be any number from 4.75 to infinity.

Answer 2

9

Answer 3

10 squared and 10+5 is 115.
115 is divisible by 5, and so is not a prime number.

Answer 4

There are infinitely many counterexamples. This is a well known theorem: If f(x) is any non constant polynomial with integer coefficients, then there are infinitely many values t such that f(t) is composite.

Here is the proof: Suppose a is any integer for which f(a) is not 1, and consider the number f(a). Now f(a + f(a)) contains
f(a) as a factor.

Answer 5

x^2 + (x + 5) = p

let x = 5
25 + 5 + 5 = 35

35 is not prime since it is divisible by 5 and 7.

Note: There are many counterexamples. This is just one.

Answer 6

5^2 + 5 + 5 = 35

35 is not prime 35 = {1, 5, 7, 35}

Answer 7

Try, 4,5,8,9,10,14... this thing falls apart easily. There are tons of counter examples. 71 numbers between 1 and 100 are counter examples. Did you even try this?

Answer 8

p(x) = x^2 + x + 5

x.....p(x)

0.....5............the 3rd prime
1.....7............the 4th prime
2.....11..........the 5th prime
3.....17..........the 7th prime
6.....47..........the 15th prime
11.....137......the 24th prime
16.....277......the 59th prime
17.....311......the 64th prime
18.....347
21.....467
23.....557
27.....761
...............
The numbers in between the ones above are counter examples, which are infinite in number.