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

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*

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