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I have been recently thinking about the squares of prime numbers "p" and found that the number of the form p^2 - 1 is divisible by 12 except the primes 2 and 3. Is my statement correct ? So , if it is correct can anyone prove how are they divisible by 12?

Note by Sriram Venkatesan
2 years ago

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p^2-1 can be written as (p-1)(p+1).2 is the only even prime number.Hence,p+1,p-1 are even.Thus,(p+1)(p-1)is divisible by 4.Also,if we take any three consecutive natural numbers,one of them is divisible by three.Here,if we take (p-1),p,(p+1),then,p is not divisible by 3 as it is prime,thus,either (p+1) or (p-1) should be divisible by 3.Hence,p^2-1 is always divisible by 12 for prime numbers 5 and above. Akshay Bodhare · 2 years ago

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@Akshay Bodhare Thanks for your proof Sriram Venkatesan · 2 years ago

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@Sriram Venkatesan It is simple.No problem. Akshay Bodhare · 2 years ago

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\[p^2 -1 =(p-1)(p+1)\]. Now all primes satisfy: \[(p\equiv1\mod6)\ or \ (p\equiv-1\mod6)\], (as the other options can be factorized) which means that in both cases we have 6|(p-1)(p+1). Also, both of \({p}\) -1 and \({p}\) +1 are divisible by two as \({p}\) is odd ( for \({p}\) > 2). \(\therefore\) 24|\(p^{2}\) -1 \(\Rightarrow\) 12|\(p^{2}\) -1 Curtis Clement · 1 year, 12 months ago

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mADE ME THINK... Vishwathiga Jayasankar · 2 years ago

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