Question: Find the minimum positive odd interger $m$ so that $(m+4\cdot41)(m^2+4^2\cdot41^2)$ is a square number.

Any suggestion will be appreciated. Thanks.

  • $\begingroup$ Rather multiply bracket i think itll get a hint of some number $\endgroup$ – Archis Welankar Feb 27 '16 at 4:53
  • $\begingroup$ Considering the case when $m=41k$ where $k$ is a positive odd integer will give you that $m=41\cdot 5=205$ is possible. $\endgroup$ – mathlove Feb 27 '16 at 6:04
  • $\begingroup$ At this point an astute computer programmer would combine this with the result below, that m-5 is divisible by 8,and check all smaller possible cases. A less astute one would check every value from 1 to 204. Either way,it gets the answer but with no insight. $\endgroup$ – DanielWainfleet Feb 27 '16 at 7:23
  • $\begingroup$ @user254665: No "astute computer programmer" is needed. This can be done very easily in a spreadsheet such as Microsoft Excel. I am often amazed that more people do not use this resource. And the question did not prohibit such answers. $\endgroup$ – Rory Daulton Feb 27 '16 at 13:36
  • $\begingroup$ @RoryDaulton. I was thinking of a friend,who is a top-notch programmer. If you tell him that an answer lies in a known bounded set his response will be, "Compute it then." I think I'll try to get him to compute Ramsey numbers :) $\endgroup$ – DanielWainfleet Feb 27 '16 at 19:51


Since $m$ is an odd positive integer, so we must have $$(m+4\cdot 41)(m^2+4^2\cdot 41^2)\equiv 1 \pmod 8$$ $$m^3+4^3\cdot 41^3 + m\cdot 4\cdot 41(m+4\cdot 41) \equiv 1 \pmod 8$$

  • $\begingroup$ So $m\equiv 5 \pmod 8$ because for odd $m$ we have, (mod $8$) $m^2\equiv 1$ and $m^3\equiv m^2.m\equiv m.$ $\endgroup$ – DanielWainfleet Feb 27 '16 at 7:13

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