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Is it possible to find the first common value of these two sequences?

$m^2 + 928m + 687: 1616, 2547, 3480, 4415, 5352, \dots$

and

$n^2 + 54n + 729: 784, 841, 900, 961, 1024, \dots$?

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    $\begingroup$ How about $m=33$, $n=153$? Both lead to $32400$ (After that, it takes a little while) $\endgroup$ – Hagen von Eitzen Apr 15 '16 at 14:06
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$$m^2+928m+687 = n^2+54n+729 \\ (m+464)^2-(n+27)^2 = 464^2-687 \\ (m+n+491)(m-n+437) = 214609 = 317*677 = 1*214609\\ \cases{m=33, n=153 \\ m=106841, n=107277}$$

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    $\begingroup$ The second $m$ should be 106841, I suppose $\endgroup$ – Hagen von Eitzen Apr 15 '16 at 14:09
  • $\begingroup$ @HagenvonEitzen Yes, it really should. Silly mistake. $\endgroup$ – Abstraction Apr 15 '16 at 14:12
  • $\begingroup$ Thanks! But isn't there a way without the need of finding factors of the number 214609? $\endgroup$ – Peter Apr 15 '16 at 14:26
  • $\begingroup$ If there would be one, you could use it to factorize $214609$ instead. So, were you to find a way to solve such general task significantly faster than by factorization, it would be a major breakthrough in cryptography. Possible, but unlikely. $\endgroup$ – Abstraction Apr 15 '16 at 14:29
  • $\begingroup$ I see! Thanks a lot. $\endgroup$ – Peter Apr 15 '16 at 14:41

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