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It seems that difference of squares of any twin primes $+1$ will always lead to
number which might be
a) A square of a twin prime
b) Itself a twin prime

$C$ = ($A^2$-$B^2$ )+$1$ ------> $(1)$

Where
$C$ --- > might be a twin prime or square of a twin prime,
$A$ and $B$ are twin primes where $A$ is > $B$

My questions is whether eqn ($1$) is true?

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When you say $A$ and $B$ are twin primes with $A > B$, do you mean $A = B + 2$, or that $A$ can be any twin prime greater than $B$? –  Michael Albanese Dec 27 '12 at 12:57

3 Answers 3

up vote 4 down vote accepted

If $A$, $B$ are twin primes, they differ by 2, so the conjecture seems to be that

if $C = 2(A + B) + 1$, with $A$, $B$ twin primes, then $C$ is either a twin prime or square of a twin prime.

That's quickly falsified by taking $A = 101, B = 103$. For then $C = 409$ which is neither a twin prime nor the square of one.

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Similarly 137, 139, 553 and many (most?) larger examples –  Henry Dec 27 '12 at 13:02

$\,(103^2-101^2)+1=409\,$ , which is neither of (a)-(b), though it is a prime.

$\,(4801^2-4799^2)+1=19201=7\cdot 13\cdot 211\,$ , which is neither of (a)-(b) and not even a prime

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Again the number 19201 is a product of primes? –  Shan Dec 27 '12 at 13:14
    
Well, any natural number is a prime or a product of primes except $\,1\,$....is this what you actually meant to ask? –  DonAntonio Dec 27 '12 at 13:15
    
I agree..need to work more on this.. –  Shan Dec 27 '12 at 13:19

Some counterexamples:

$$ \begin{array}{c|c} p & (p+2)^2-p^2+1 \\\hline 137 & 7^1 \times 79^1 \\ 179 & 7^1 \times 103^1 \\ 197 & 13 \times 61^1 \\ 227 & 11^1\times 83^1 \\ 269 & 23^1\times 47^1 \\ 347 & 7^1\times 199^1 \\ 431 & 7^1\times 13^1\times 19^1 \end{array} $$

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