Puzzle 811. Primes and Sum of consecutive triangulars-II

I use the following notations :

T(n)=n(n+1)/2 = the nth triangular number

U(n) = T(n)+T(n+1)+T(n+2) = (3n^2+8n+9)/2 (derived in puzzle 810)

V(m) = T(m)+T(m+1)+T(m+2)+T(m+3)+T(m+4)+T(m+5) = 3m^2+18m+35 (id).

Q1.

We should solve the equation U(n) +1= V(m) or :

(3n^2+8n+9)/2 + 1 = 3m^2+18m+35

This is equivalent to

n^2 + 3n - (2m^2 + 12m +20) = 0

We can consider this as a quadratic equation in the unknown n.

This will have integer solutions iff its discriminant (which is 9 + 4(2m^2 + 12m +20)) is the square of an integer x.

Thus, we should solve the equation :

x^2 = 8m^2 + 48m + 89 (*)

and then put n = (x-3)/2. This leads us very quickly to the solutions :

m = 5, 10, 44, 73, 271, 440, 1594, 2579, 9305, 15046, 54248, 87709, 316195, 511220, ...

n = 10, 17, 65, 106, 386, 625, 2257, 3650, 13162, 21281, 76721, 124042, 447170, 722977, ...

The corresponding pairs (U(n),V(m)) of consecutive integers are :

(199, 200), (514, 515), (6634, 6635), (17335, 17336), (225235, 225236), (588754, 588755), (7651234, 7651235), (20000179, 20000180), (259916599, 259916600), (679417210, 679417211), (8829513010, 8829513011), (23080184839, 23080184840), (299943525619, 299943525620), (784046867194, 784046867195), ...

Though there are many solutions, this does not give answer whether or not there are infinitley many ones.

But, we can rewrite (*) in the form x^2 - (8m^2 + 48 m +72) = 17 or :

x^2 - 2y^2 = 17 (**)

with y = 2m + 6.

To solve (**) we can use the well known theory of numbers of the form a+b*Sqrt(2) (a, b integers).

There are two 'small' solutions (x,y) = (5,2) and (x,y) = (7,4).

From these solutions we can obtain an infinitude of others by the following rule :

if (x,y) is a solution, then so is (3x+4y,2x+3y)

This let us find all the solutions.

Here, we should solve the equation V(m)+2 = U(n) or :

n^2 + 3n - (2m^2 + 12m +22) = 0

Here also, this is solvable in integers provided there is an integer x such that :

x^2 = 8m^2 + 48m + 97 (***)

and n = (x - 3)/2.

This leads us quckly to the solutions :

m = 2, 27, 172, 1017, 5942, 34647, 201952, ...

n = 6, 41, 246, 1441, 8406, 49001, 285606, ...

The corresponding pairs (V(m), U(n)) are :

(83, 85), (2708, 2710), (91883, 91885), (3121208, 3121210), (106029083, 106029085), (3601867508, 3601867510), (122357466083, 122357466085), ...

If we work as in Q1, we should first solve

x^2 - 2 y^2 = 25,

which has the solution (x,y) = (5, 0) and all other solutions come from the rule that (3x+4y,2x+3y) is also a solution.

So, in every solution x and y are seemingly divisible by 5. Since n = (x - 3)/2 this implies that n allways will be congruent to 1 mod 5.

Thus, U(n) = (3n^2 +8n + 9)/2 = 0 mod 5.

So, there is no twinprime pair solution !

N.B. The rule to be applied on (m,n) is (in both cases Q1 and Q2) :

if (m,n) is a solution then so is (3m+2n+9,4m+3n+15).