If p is a prime number, what is the value of p ? (1) \sqrt

I wrote this question, so below is my solution:

(1) \(\sqrt[3]{1-p^2}=-n\), where \(n\) is a prime number.

It's clear that \(p\neq{2}\) as in this case \(\sqrt[3]{1-p^2}\neq{integer}\), so \(p\) is a prime more than 2, so odd. Then \(1-p^2=odd-odd=even\) --> cube root from even is either an even integer or not an integer at all, we are told that \(\sqrt[3]{1-p^2}=-n=integer\), so \(-n=even=-prime\) --> \(n=2\) (the only even prime) --> \(\sqrt[3]{1-p^2}=-2\) --> \(p=3\). Sufficient.

(2) \(8p^2+1=m\), where \(m\) is a prime number.

\(p\) can not be any prime but 3 for \(8p^2+1\) to be a prime number: because if \(p\) is not 3, then it's some other prime not divisible by 3, but in this case \(p^2\) (square of a not multiple of 3) will yield the remainder of 1 when divided by 3 (not a multiple of 3 when squared yields remainder of 1 when divided by 3), so \(p^2\) can be expressed as \(3k+1\) --> \(8p^2+1=8(3k+1)+1=24k+9=3(8k+3)\) --> so \(8p^2+1\) becomes multiple of 3, so it can not be a prime. So \(p=3\) and in this case \(8p^2+1=73=m=prime\). Sufficient.

Answer: D.
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A is ans
on solving a we get
1+ n^3 = P^2
so it will be satisfied with n=2 and p=3..
the reason is..p will be alws greater than n..so p will alws be odd no..so its square will be alws odd...
so n^3 must be even which is possible with only prime no 2..
so 1+n^3 will bcome odd...
so satisfied...

2) on solving eq bcom 8p^2 +1 =m..so I m unable to find any prime no which is 24k+1..where k is any + integer.. so not satisfied

(1) \(\sqrt[3]{1-p^2}=-n\) OR \(p^2=1+n^3=(1+n)(n^2-n+1)\)
If n=2, p=3
If n>2, then n is odd, (n+1) is even, hence p^2 is even, but this is only possible if p=2 and if p=2, n is cube_root(-3) which is not an integer
So only possibility is p=3
Sufficient

(2) \(8p^2+1=m\)
p is an integer and a prime so it can either be of the form 3k+1 or 3k-1 or be 3 (all other numbers of form 3k are composite)
8*(3k+1)^2+1=8*(9k^2+6k+1)+1=3*(24k^2+16k+3) .. which cannot be a prime since k>=1
8*(3k-1)^2+1=8*(9k^2-6k+1)+1=3*(24k^2-16k+3) .. which cannot be a prime since k>=1
8*3^2+1=73 is a prime
So only possible value of p is 3
Sufficient

Answer is (D)
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hey Man
p is an integer and a prime so it can either be of the form 6k+1 or 6k-1... not 3k+1 and 3k-1

You can always divide all integers into 3 sets : 3k , 3k+1 , 3k-1
In case of primes, it is easy to see only one element in the first set, i.e., 3
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ya agreed D is ans..Thnx

Statement 2: take prime = 6k+1 and 6k-1

\(8*((6k+1)^2 + 1) = 8*36k^2 + 12k +9 =\) 3* something => not a prime
\(8*((6k-1)^2 + 1) = 8*36k^2 - 12k +9 = 3*\) something => not a prime

so any prime number of the form 6k+1 and 6k-1 is not possible.
lowest prime of the above form is for k=1 i.e. 5.

Thus only 2 possibilities p=2 and p=3. Only p=3 satisfies.

BUNUEL- Great question.
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great question Bunuel

I am really confused. Can someone please explain me,

Statement 2:

I understand that if we work out the
8P^2 + 1 : where P = 2
we get m = 33, which is NOT a prime.

Great! I am on-board till here.

Is the strategy to try every prime number from 2, onwards??

How is the value of P chosen??

You state P cannot be any prime but 3 for 8P^2 + 1 to be a prime number. How can you be so sure? I do not have an exhaustive list of all prime number, but this claim is little far fetched.

Please clear my confusion.

There are infinitely many prime numbers so there is no "list of all primes".

Next, \(p\) can not be any prime but 3 for \(8p^2+1\) to be a prime number: because if \(p\) is not 3, then it's some other prime not divisible by 3, but in this case \(p^2\) (square of a not multiple of 3) will yield the remainder of 1 when divided by 3 (not a multiple of 3 when squared yields remainder of 1 when divided by 3), so \(p^2\) can be expressed as \(3k+1\) --> \(8p^2+1=8(3k+1)+1=24k+9=3(8k+3)\) --> so \(8p^2+1\) becomes multiple of 3, so it can not be a prime. So \(p=3\) and in this case \(8p^2+1=73=m=prime\).
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Hi Bunuel,

OK.
But I am still not clear about why is it important to be divisible by 3.

"because if \(p\) is not 3, then it's some other prime not divisible by 3, but in this case \(p^2\) (square of a not multiple of 3) will yield the remainder of 1 when divided by 3 (not a multiple of 3 when squared yields remainder of 1 when divided by 3)"





I guess, you will have to explain this a little more.

Because if \(p\) IS NOT divisible by 3 then \(8p^2+1\) IS divisible by 3 and thus can not be a prime, so \(p\) must be divisible by 3, only prime divisible by 3 is 3.
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Yes nice question- 700+
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D ...intially i thought it as A coz i was wrongly marked 57 as prime Silly mistake :0

Tough one, I could solve the question by plugging in the numbers but didn't think of trying (especially for 2nd statement) the above concepts, despite knowing for a number to be prime it should be either \(6k+1\) or \(6k-1\). Similarly, for statemet1 despite knowing that \(p\) can not be \(2\) and it would be more than didn't think that it must be odd and apply Odd & Even concept here.

Thank you! for reminding me to use concepts rather than just guessing and then getting doubtful solutions (and doing silly mistakes).

Please find the solution as attached.

Answer: option D

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Dear GMATinsight,

Thanks for you explanation. I did the same as you did in solving those question. But I had a doubt that there maybe a higher prime number that satisfy fact 2 so I kept doing this to 13 & 17. How should I be confident to stop at 11 like you?

Mo2men

To be Confident

1) You need to find out a pattern in solutions because there are always patterns followed among all acceptable or unacceptable solutions

2) To find a pattern you need three instances/examples to observe

I saw the pattern in solutions that each of the solutions that I was finding was a multiple of 3 so I concluded that this pattern is going to continue

Trusting the pattern based on three instances usually gets you the correct answer in 99% cases and the remaining 1% are usually made by Manhattan intentionally to prove this thought process incorrect. Thankfully Manhattan doesn't make GMAT so the chances are bleak that you get wrong answer trusting the pattern
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VeritasKarishma

To express primes >3, we use 6n+1 and 6n-1

If we use 3k+1, wouldn't it then erroneously include 4 when k=1

Kindly clarify!