Problem of simultaneous primes

Any number p can be congruent as 0,1,2,3,4 to (mod 5) and so we use congruence in p.

any square number can be congruent to (mod 5) as 0,1,4 and taking all that as basis we proceed to put congruence in p^2 in the below cases.

Case1 - CONGRUENT TO 0

p can obviously be congruent to 0 and also be prime only at p=5

checking other forms we get 4p^2+1=101 and 6p^2+1=151 each being prime.

therefore p=5 is a solution.

Case2 -CONGRUENT TO 1

p congruent to 1 (mod 5) but then 4p^2+1 is congruent to 0 (mod 5) and being divsible by 5.

also 4p^2+1 cannot be 5 for any integer.(primes are always positive integers)

so p is not congruent to 1 (mod 5)

Case3 - CONGRUENT TO 2

p congruent to 2 would mean that 4p^2+1 is congruent to 0 (mod 5)

and so above deduction it s not possible for any integer.

so p is not congruent to 2 (mod 5)

Case4 - CONGRUENT TO 3

p congruent to 3 would mean that 6p^2+1 is congruent to 0 (mod 5) this implies for no solution for p = 3 (mod 5)

so p is not congruent to 3 (mod 5)

Case5 - CONGRUENT TO 4

p congruent to 4 would mean that 4p^2+1 is congruent to 0 (mod 5) this implies for no solution for p = 4 (mod 5)

so p is not congruent to 4 (mod 5)

therefore only solution to this problem is p=5 answer =5