IIT JEE , AIEEE Forum - Mathematics , Algebra - Find all values of n, for which n, 4n2 +1, 6n2 +1 are all prime numbers?

abcd86

Find all values of n, for which n, 4n²+1, 6n² +1 are all prime numbers?

b_srikalyan009

except for n=5 no other value satisfies the condition.

we can prove it logically...

n should be a prime number hence it should be odd(2 is not included).

let the prime number have 1 in units plase then 4n^2+1 will have 5 in the units place hence 1 is not permitted.

let the prime number have 3 in units place then 6n^2+1 will have 5 in units place hence it is not permitted.

similarly for 7,9 also it is not possible.

but only when n=5 it satisfies the condition

spideyunlimited

prime number is odd?!!!! who u kidding man

n = 2 is even n prime

coool_shetty

it was a math olympiad question last year ...
actually the question was to find an odd number which satisfied the properties given..
just take a number as
i)5x
ii)5x+1
iii)5x+3.
when u substitute the props ,only 5x satisfies .so number is 5!!!
please rate!!!!

aditya_arora04

Please, if anyone has a proper ( understandable ) solution, please help !!!!

b_srikalyan009

@spydyunlimited
i mentioned that 2 is not included

spideyunlimited

units place of squares of all numbers end in one out of
0,1,4,9,6,5

for n to be a prime number , it cant have 0, 2, 4, 6, 8, 5 in units place ( except for n = 2 and 5)
thus its square cant end with 0, 4, 6, or 5 (except for n = 2.. as 2 is the only even prime number and n = 5 as mentioned in previous line)

thus we are left with 1 and 9 .. n square can only end with 1 and 9
and when we have this condition we see that for any such value of n we always have one composite number (non prime) out of 4n^2 + 1 and 6n^2 + 1
because with last digit as 1, 4n^2 + 1 will have 5 as units place
and with last digit as 9, 4n^2 + 1 will have 5 in units place

thus we are only left with n = 2 and n = 5.

but at n = 2 , 6n^2 + 1 becomes 25 which is not prime
thus the only solution is n = 5

*PLZ RATE*

priyesh

Excellent solution gaurav superb!!!!!!!!!!!!!!!!!
rating u dude

spideyunlimited

ur online at this time priyesh!! :)

metal

Good solution Srikalyan

metal

Good solution spidey

vigi

the nos. can end in 1,3,5,7,9.
by trial & error, we can rule out nos. ending in 1,3,7,9.
now we have nos. ending in 5.
but every no. other than 5 is not prime(it is a multiple of 5)
So, n=5.

metal

We know the answers already, don't we????

Spidey and Srikalyan have already given the answers.

So, why the hell are others going on giving the answers ???????

priyesh

very rightly said by metal

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