candidate = -2+p*2p5
Is candidate prime?
Example p=53 so 2p5 =101
5351=-2+53*101
5351 is prime
Next try p=47 so 2p5 = 89
4181=-2+47*89
4181=37*113
Next try p=59 so 2p5 = 113
6665=-2+59*113
6665=5*31*43
Next try p=61 so 2p5 = 117
7135=-2+61*117
7135=5*1427
Next try 67 so 2p5 = 129
8641 is prime
Now algebraically we might say -2+p*(2p-5) which is rewritten 2p^2-5p-2
Next try 71 so 2p5 = 137
9725=-2+71*137
9725=5*5*389
Next try 73 so 2p5 = 141
10291=-2+73*141
10291=41*251
Next try 79 so 2p5 = 153
12085=-2+79*153
12085=5*2417
Next try 83 so 2p5 = 161
13361=-2+83*161
13361=31*431
Next try 89 so 2p5 = 173
15395=-2+89*173
15395=5*3079
Next try 97 so 2p5 = 189
18331=-2+97*189
18331=23*797
Next try 101 so 2p5 = 197
=-2+101*197
=5*23*173
Next try 103 so 2p5 = 201
20701=-2+103*201
20701=127*163 is composite
Next try 107 so 2p5 = 209
22361=-2+107*209
22361=59*379 is composite
Next try 109 so 2p5 = 213
23215=-2+109*213
23215=5*4643 is 5 divisible
Next try 113 so 2p5 = 221
24971=-2+113*221
24971 is prime
Answer will be divisible by 5 when -2+2p^2 is divisible by 5
( or written another way answer cannot be divisible by 5 when -2+2p^2 is not divisible by 5 )
Now we redo the above example but with slightly different assignment for p
Set the prime 113 then deduct a 100 and use that figure as p
13 is prime
p=13
113=p+10^2
221 = p*(p+4)
-2+(p+(2*5)^2)*p*(p+4)
prime 24971==-2+113*221 ( 221 is 13*17 )
p=17
-2+(p+(2*5)^2)*p*(p+4)
41767=-2+117*17*21
41767=11*3797
p=19
-2+(p+(2*5)^2)*p*(p+4)
51201=-2+(7*17)*(19*23)
51201=149*349
p=1053 so p+100 is 1053
p2=1053
prime 1283313211==-2+1153*(1053*1057)
( 1283313213==(3^4)*7*13*151*1153 )
Now let us vary things slightly by defining p2=p+100 and having our answer derived as -2+(p2^2)*(p2+4)
p=953 which is prime and then p+100 is 1053
p2=1053
prime 1172011111==-2+1053*(1053*1057)
( 1172011113==(3^8)*7*(13^2)*151 )
