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If p is a prime less than 500, how many number of the form 3p + 1 are perfect squares?
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Anybody??????
One.
\[ p=5; 3p+1 = 16 = 4^2 \]
A perfect square ends with 00,4,5,6,9 so p must have its unit digit as 1 or 5 and then i substituted the value
\(3p+1 = k^2 \) (say) \[ 3p = (k+1)(k-1) \] Now there will be two cases either (k+1)=3 and (k-1) = p or viceversa.
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can you do the rest?
Anyways, If \( (k+1)=3 \) and \( (k-1) = p \implies p=1 \) which is not prime If \( (k+1)=p \) and \( (k-1) = 3 \implies p=5 \implies 3p+1 = 16 = 4^2 \) There will be only one such prime. Q.E.D
Oh yes
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