# Find $m$ so that $(m+4\cdot41)(m^2+4^2\cdot41^2)$ is a square number

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.

• Rather multiply bracket i think itll get a hint of some number – Archis Welankar Feb 27 '16 at 4:53
• Considering the case when $m=41k$ where $k$ is a positive odd integer will give you that $m=41\cdot 5=205$ is possible. – mathlove Feb 27 '16 at 6:04
• 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. – DanielWainfleet Feb 27 '16 at 7:23
• @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. – Rory Daulton Feb 27 '16 at 13:36
• @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 :) – DanielWainfleet Feb 27 '16 at 19:51

Since $m$ is an odd positive integer, so we must have $$(m+4\cdot 41)(m^2+4^2\cdot 41^2)\equiv 1 \pmod 8$$ $$m^3+4^3\cdot 41^3 + m\cdot 4\cdot 41(m+4\cdot 41) \equiv 1 \pmod 8$$
• So $m\equiv 5 \pmod 8$ because for odd $m$ we have, (mod $8$) $m^2\equiv 1$ and $m^3\equiv m^2.m\equiv m.$ – DanielWainfleet Feb 27 '16 at 7:13