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If $p_1,p_2$ are odd prime numbers , is it possible that

$\dfrac{p_1p_2-1}{p_1+p_2}$ is odd natural number greater than 1.

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Hint: Suppose $\dfrac{p_1p_2-1}{p_1+p_2} = 2k+1$ for some positive integer $k$. Then, we get:

$p_1p_2-1 = (2k+1)(p_1+p_2)$

$p_1p_2-(2k+1)p_1-(2k+1)p_2-1 = 0$

$p_1p_2-(2k+1)p_1-(2k+1)p_2+(2k+1)^2 = (2k+1)^2+1$

$(p_1-(2k+1))(p_2-(2k+1)) = 2(2k^2+2k+1)$

Clearly, $2k^2+2k+1$ is odd, so exactly one of $p_1-(2k+1)$ and $p_2-(2k+1)$ can be even.

Do you see how this helps?

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  • $\begingroup$ Yeah , i got it !! So , not possible . $\endgroup$ – hanugm Mar 16 '15 at 6:39
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Exactly one of $p1+p2,p1p2-1$ is a multiple of 4.

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Write the odd numbers $p_1,p_2$ and the result als numbers of the form $2k+1,2j+1,2m+1$ Then $$ {(2k+1)(2j+1)-1 \over 2k+1+2j+1} = 2m+1 \tag 1$$ Expanded
$$ {4kj + 2(k+j) \over 2k+2j+2} = 2m+1 \\ 2kj + (k+j) = (2m+1)(k+j+1) \\ 2kj + (k+j) = (2m+1)(k+j)+(2m+1) \\ 2kj = 2m(k+j)+2m+1 \\ 2kj = 2m(k+j+1)+1 $$

and finally $$ 2(kj-m(k+j+1)) = 1 \tag 2 $$ In the last row, the lhs is even and the rhs is odd, so no solution possible.

Note,that the property of $p_1,p_2$ being prime was not used, only that they are odd numbers

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$$\dfrac{p_1p_2-1}{p_1+p_2}=p_1-\left(\dfrac{p_1^2+1}{p_1+p_2}\right)$$ Therefore it is enough to find primes $p_1, p_2$ such that $$p_1^2+1|p_1+p_2.$$

One such way is, primes of the form $$p_2=p_1^2-p_1+1.$$

$(3,7),(7,43),(13,157)$ are the smallest such pairs.
Also I have no idea about the number of such prime pairs.

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  • 2
    $\begingroup$ But how can with that examples the result be an odd number (as required in the OP)? $\endgroup$ – Gottfried Helms Mar 16 '15 at 8:03
  • $\begingroup$ @GottfriedHelms: Thank you. I got it. You have gave a nice solution. Any way I will try to gave another proof. $\endgroup$ – Bumblebee May 12 '15 at 9:32

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