StevenXClontz’s Twitter Archive—№ 8,116

@dmarain @kennedy_tierney @edubabbIe @mpershan @mathinct678 @benjamin_leis @MrCorleyMath @normalsubgroup @mATH_e_matics @HarMath @johnjoy1966 @jamestanton @neiltyson @MoMath1 @RaminKhosravi4 @Mathgarden @MathTechCoach The boring answer lol n⁴+6n³+11n²+6n+1 = (n²+3n+1)² Less boring: I factored this by noting that I needed 2→25=5² and it must be of the form (n²+kn+1)² for some k, yielding k=3.