I would like to call the subject of this page Biquadratic Paired Primes (BQPPs). However, since biquadratic is another term for quartic, or fourth degree polynomials, I am being cautious about it. (Well. like biweekly, it could be fairly used to mean two completely different things....)
The sense in which I'm using biquadratic is the sense of every other quadratic interval: in fact, a span that bridges one complete interval and two partial intervals. Odd Squared Paired Primes is not entirely accurate either, as odd square implies the odd perfect squares only. In fact, the pairs of primes I'm going to describe can span two odd perfect squares (with an even one between) or two even perfect squares (with an odd one between).
A simple definition would be this:
Two primes that are separated by the equivalent of 2 quadratic intervals. Both primes are greater than their preceding perfect squares by the same amount, or offset. The respective perfect squares can be both odd, in which case the offset is even, or both even, in which case the offset is odd.
The following diagram illustrates the relationship for two pairs of biquadratic primes:
Notice the common difference, or offset, between the primes of each pair and their preceding perfect squares: 23 and 43 have an odd offset of 7 from even perfect squares 16 and 36; 29 and 53 have an even offset of 4 from perfect squares 25 and 49.
In this example, 7 and 19 are offset by 3 from the even perfect squares 4 and 16:
Interval 1: 2*2 = 4 + 3 = 7
Interval 2: 3*3 = 9 + 3 = 12
Interval 3: 4*4 = 16 + 3 = 19
Another oddoffset one: 71 and 107 are offset by 7 from 64 and 100:
Interval 1: 8*8 = 64 + 7 = 71
Interval 2: 9*9 = 81 + 7 = 88
Interval 3: 10*10 = 100 + 7 = 107
The even offsets work the same way, but from the odd perfect squares. For example, 97 and 137 are offset by 16 from 81 and 121:
Interval 1: 9*9 = 81 + 16 = 97
Interval 2: 10*10 = 100 + 16 = 116
Interval 3: 11*11 = 121 + 16 = 137
In some sense, these paired primes may be related to twin primes. That's pure conjecture, just an impression based on looking at their frequent colocation. In these examples, one or both paired primes are also twin primes:
5  7  9
17  19
71  73
107  109
137  139
Paired Primes 
Perfect Squares 
Offset 
Product 
3 , 11 
1 , 9 
2 
33 
5 , 17 
4 , 16 
1 
85 
7 , 19 
4 , 16 
3 
133 
13 , 29 
9 , 25 
4 
377 
17 , 37 
16 , 36 
1 
629 
23 , 43 
16 , 36 
7 
989 
29 , 53 
25 , 49 
4 
1537 
43 , 71 
36 , 64 
7 
3053 
67 , 103 
64 , 100 
3 
6901 
71 , 107 
64 , 100 
7 
7597 
73 , 109 
64 , 100 
9 
7957 
97 , 137 
81 , 121 
16 
13289 
The first 1,895 are available for download in CSV format (see left margin).
As you can see, paired primes are the rule rather than the exception at the beginning of the number line. There are 32 primes, up to and including 137  and 24 of them are paired. So which prime numbers less than 137 are not paired?
2, 31, 41, 61, 79, 83, 89, 101, 127
After the highdensity start, the numbers fall off steeply  in fact, there are just 25 more pairparticipating primes up to 1,000. That's 49 paired primes out of 168 primes altogether (about 30%). By 100,000, it appears that 1,895 of the 9,592 primes are paired. That's about 20%.
It seems that the first prime of every biquadratic pair is less than the second prime of the preceding pair. In other words, the pairs appear consistently to overlap. This tandem or relay relationship is shown here (and in the illustration above).
131 , 179 , 121 to 169 , 10 , 23449
173 , 229 , 169 to 225 , 4 , 39617
197 , 257 , 196 to 256 , 1 , 50629
211 , 271 , 196 to 256 , 15 , 57181
223 , 283 , 196 to 256 , 27 , 63109
229 , 293 , 225 to 289 , 4 , 67097
I haven't looked that hard for an exception yet. There may well be exceptions in the prime pairs under 100,000 if you care to look at this text file (101KB). (You could actually extend this claim to the first pair, 3 and 11, if you are willing to shed your inhibitions and allow 0 to be a perfect square and, dare I say, pretend 1 is a prime... confusing since it's a perfect square also... OK let's just forget that.)
There may be a PhD awaiting a mathematician here....
A pronic number is near the heart of every biquadratic pair. Notice that the "midpoint" in the paired prime examples, occupying the intermediate quadratic interval, was a pronic number in the first example (3*3+3 = 12). This will only occur when the offset equals the root of the pronic number. So, for example if the offset is 7, the pronic would have to be 56 (7*7+7).
I think the following additional assertions can be made about biquadratic pairs and pronic numbers:
 Every biquadratic pair must, by definition, contain within its range an absolute pronic number (in the intermediate interval).
 The difference between the primes of every pair that contains within its range the same absolute pronic must be the same.
Speculatively,
The number of quadratic intervals that occur before a relative pronic number (the pairedprime difference in point 2) occurs appears to increase by a finite difference of 1.
I would love to be able to say that every absolute pronic number has a BQPP, but this is not the case. About 1 in 5 do not, and this frequency of nonBQPP pronics gradually increases.
Before you start your thesis, let me check my data...!
The products of BQPPs could be called "odd squared composites". Interestingly, they bear a passing resemblance to primes  although they are mostly not pseudoprimes. (FLT correctly identifies most of them to be composites.) The resemblance is the superficial one that the last digit is always 1, 3, 7, or 9. There is one single exception to this  85, the product of the second prime pair (comprising 5, of course, and 17).
For what it's worth, these composites are the squared median of the primes minus 2 plus the offset multiplied by 4. If you have something against straight multiplication, you can arrive at the answer this way:
(P1 + P2)/2 1 ^ 2 + offset * 4
For example: 113 * 157 = 17741
(113 + 157)/2 = 135
135  2 = 133
133 ^ 2 = 17689
17689 + (13 * 4) = 17741
Now this suggests doing a reverse calculation from any given number  a trivial calculation for even very large numbers  as an initial step in factoring it, to see if it belongs to this sizeable class of composites. You would do it like this:
Say you've been given a big number that looks like it might be a prime  that is, it ends in 1, 3, 7, or 9. (You can pretend that it's much bigger than this one if you would like.)
N = 878199517529
You would proceed as follows:
 Find the square root of N
878199517529 ^ 1/2 = 937123.0002134191562657475399381
 Convert the result to an integer and square it to get the nearest perfect square to N
937123 ^ 2 = 878199517129
 Now subtract this perfect square from N and divide the result by 4
878199517529  878199517129 = 400
400 / 4 = 100
This is the offset if N is the product of paired primes.
<<If the number is not even, you do not need to proceed  N cannot be the product of paired primes.>>
Now you need to determine the perfect squares that straddle 937123. The easiest way to do this is to find its square root  or the quartic root of N (also called the biquadratic root... yes, I can use this term without guilt!).
 Find the quartic root of N
878199517529 ^ 1/4 = 968.051135123253
 Convert the result to an integer, then subtract 1 to get the root of the lower perfect square and add 1 to get the root of the higher perfect square. Then square these roots:
968 = (967,969)
967 ^ 2 = 935089
969 ^ 2 = 938961
 All that remains is to add the offset of 100 to each of these numbers:
935089 + 100 = 935189
938961+ 100 = 939061
 We're done. Let's check that the factors are correct:
935189 * 939061 = 878199517529
We have shown that a large number N that is potentially a prime is in fact a composite of biquadratic paired primes. However, this is not a proof unless the factors are prime factors. Unfortunately, this procedure does not prove that. Try a demonstration of the "algorithm", exactly as described, in JavaScript.
Biquadratic paired primes can be easily visualized using a grid with an axis of perfect squares and an axis of offsets from the perfect squares. The following illustrates the most salient points. You can generate this in Excel using the VBA code below.
Notes:
 Only the subset of paired primes for even perfect squares with odd offsets are demonstrated.
 Because of the 256column limitation for an Excel worksheet, data in rows after perfect square 484 are not complete.
 Only twin primes that are also paired primes are outlined (in red).
Copy and paste into a VBAProject ThisWorkbook object.
Michael M. Ross
Please let me know if you've seen this relationship or an analogous one described elsewhere.
