5 2 = 2 5 = 2 4 + 1 1 9 2 = 3 6 1 = 2 4 × 1 5 + 1
We note that 5 and 19 in the equations above are primes. Can we generalize that for any prime p ≥ 5 , p 2 = 2 4 n + 1 , where n is an positive integer?
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p 2 − 1 = ( p − 1 ) ( p + 1 )
Now, ( p − 1 ) , ( p + 1 ) are two numbers that lie before and after p respectively.
Since, p is a prime, it has no factors other than 1 and itself.
Since, ( p − 1 ) , p and ( p + 1 ) are three consecutive integers and p is odd, both ( p − 1 ) and. ( p + 1 ) are even i.e, have 2 as a factor.
Now, since p ≥ 5 either of ( p − 1 ) or ( p + 1 ) have 4 as a factor.
Since, ( p − 1 ) , p and ( p + 1 ) are three consecutive integers , one of them has 3 as their factor. Since, p is prime, the other two integers have 3 as a factor.
Hence, we can conclude that, ( p − 1 ) ( p + 1 ) has 2 , 3 , 4 as its factor.
Therefore, ( p − 1 ) ( p + 1 ) = ( 2 × 4 × 3 ) n = 2 4 × n ⇒ p 2 = 2 4 × n + 1
Hence, Y e s it holds true for any prime greater than 5.
N o t e : This holds true even for 2 and 3 . But here n is not an integer.
2 2 = 4 = 3 + 1 = 2 4 × 8 1 + 1
3 2 = 9 = 8 + 1 = 2 4 × 3 1 + 1
@Niraj Sawant , I have amended your question wording. You don't need a comma after "where". You should not capitalize "where" because it is not a start of a sentence. I have changed p to lowercase because it is usually the case for prime. Wondering why you so many spaces before and after LaTex code. However many spaces you enter it will only show 1 space. Just enter 1 space will do.
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Can you change my question?? I wasnt knowing that. Anyways, thanks. Usually what happens when I dont put much spaces between Latex and Non-Latex, the Latex part and the Non-Latex part next to it overlap. I dont know why but it happens. Not all times, but sometimes.
Furthermore, a doubt how to calculate ( 1 / 4 ) ! without calculator. Any help would be greatly appreciated.
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I am a moderator. I can edit problems. I didn't know about the overlap problem.
( 4 1 ) ! = Γ ( 1 + 4 1 ) = Γ ( 4 5 ) , where Γ ( ⋅ ) denotes the gamma function (check this link). You may use calculus to solve it but it may be very difficult. You can use Wolfram Alpha (check this link ). It is free.
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@Chew-Seong Cheong – Yeah, I know about Gamma Function. Actually, I wanted to post a problem on 2 . 5 ! ! to my sequel of 2 . 5 ! .
So, I needed a more generalized solution which doesnt use any softwares. Can you do me a favour? Can you please send me screenshots of method of solving it as given in Wolfram Alpha to my email. My email: nirajsawant2313@gmail.com
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@A Former Brilliant Member – I think I saw the problem and found that the answer was wrong. From Wolfram Alpha check this link . Can you see the link? For double factorial it is not as simple.
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@Chew-Seong Cheong – Yeah, I deleted the double factorial problem. No one was solving it neither I got any reports. I checked out your link, but it just displays the formula but to see the method, I need to have a pro version. That's what I was telling that in case you have the pro version, you can take screenshots of that method and email me.
My answer turned out to be 2.150.... Which isnt exact but close, is solving double factorial of fractions really that tough?!
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We note that all primes p ≥ 5 , are odd numbers indivisible by 3. We can express any odd number larger than 3 which is indivisible by 3 as 6 k ± 1 , where k is a positive integer. Then we have:
( 6 k ± 1 ) 2 = 3 6 k 2 ± 1 2 k + 1 = 1 2 k ( 3 k ± 1 ) + 1
Since when k is odd, 3 k ± 1 is even and when k is even, 3 k ± 1 is odd, k ( 3 k ± 1 ) is always even and we can express it as 2 n , where n is a positive integer. Therefore, ( 6 k ± 1 ) 2 = 2 4 n + 1 . Yes , the square of any prime larger than 3 can be expressed as 2 4 n + 1 .