## Quant Question Of The Day: 42

[latexpage]

# Numbers

If p is a prime such that (p – 1)/4 and (p + 1)/2 are primes, then how many values of p are possible.

i) 0

ii) 1

iii) 2

iv) 3

v) More than 3

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[latexpage]

If p is a prime such that (p – 1)/4 and (p + 1)/2 are primes, then how many values of p are possible.

i) 0

ii) 1

iii) 2

iv) 3

v) More than 3

Since both (p – 1)/4 and (p + 1)/2 are integers, we can say that

p is of form 4k + 1

=> (p – 1)/4 = k

and (p + 1)/2 = 2k + 1

So, all k, 2k + 1 and 4k + 1 should be prime

If k is prime greater than 3, then

Case (i): k = 6a + 1

2k + 1 = 12a + 3 (not prime)

Case (ii): k = 6a – 1

4k + 1 = 24a – 3 (not a prime)

So, k can only be 2 or 3

k = 2

2k + 1 = 5

4k + 1 = 9 (not possible)

k = 3

2k + 1 = 7

4k + 1 = 13

So, only p = 13 satisfies

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5

Hello Arushi

5 values or p = 5 ?

More than 3

Hello Priyanka,

What was your approach?

0

What approach did you use, Samyak?

0

Hello Parth

What was your approach?

1

Correct! What approach did you use?

iii) 2

Hello Manish,

Which two values?

0

Sir, I used the fact that all the prime numbers are of the form 6n+1 or 6n-1. I replaced the value of p with the above two values and I found no solution.

1

I used hit and trial method and only 13 is that value of p that gives the other two primes .

is there any other solution ?