If you know what I mean

Both d = 7 and d = 11 satisfy the equations, albeit in the opposite way:

Take a = 19, b = 17, c = 13 (reverse consecutive primes, in that order)... then d = 7. Take a = 37, b = 41, c = 43 (consecutive primes, in that order) ... then d = 11.

We can solve the equations to get,

6a = d^2 + 9d + 2.

6b = 2d^2 + 4.

6c = 3d^2 - 9d - 6.

From a + c + 2 = 2b, we conclude that either both a & c are odd or both are even; but since a, b, c are consecutive primes, neither of them can be 2 (an even prime).

Further, a + b + c = d^2... we conclude that d > 2. So, d is an odd prime.

Finally, if d is an odd prime, it can be written as

d = 4k +/- 1, where k is a natural number.

Substituting,

6a = 16k^2 +/- 8k + 36k + 3 +/- 9

But, if d is of the form (4k + 1), then RHS is divisible by 4. This will imply, a is even. A contradiction.

Thus, d is a prime of the form: d = 4k - 1.

Thus, we can rule out 17 & 13.

Now, we can put in the values d = 7, and d = 11. We find they both give consecutive primes (in straight order or reverse order).

I saw those numbers and plugged them in, correct me if I'm wrong but they DON'T work for the last part. The closest that I got was 41 being B and then adding 43 and 37 + 1 but that only equals 81, not 82

oh sorry robert I am sorry whole heartedly due to some typing mistake the last equation went wrong the right equation is a + c+2 =2b I have edited it!!!

If you have made this question I ask you to solve my other question (got it... or not) and (heart touching isn't it)

a=19, b=17, c=13 and d =7 a+b+c=d^2 a+b-c= 3d+2 a+c+2 = 2b

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