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Question-43749




Question Number 43749 by naka3546 last updated on 15/Sep/18
Answered by MrW3 last updated on 16/Sep/18
2(n^2 +3n−1993)=k^2   k must be even ⇒k=2m  n^2 +3n−1993=2m^2   n(n+3)=2m^2 +1993  LHS is always even, even×odd=even  RHS is always odd, even+odd=odd  ⇒no solution for n!
$$\mathrm{2}\left({n}^{\mathrm{2}} +\mathrm{3}{n}−\mathrm{1993}\right)={k}^{\mathrm{2}} \\ $$$${k}\:{must}\:{be}\:{even}\:\Rightarrow{k}=\mathrm{2}{m} \\ $$$${n}^{\mathrm{2}} +\mathrm{3}{n}−\mathrm{1993}=\mathrm{2}{m}^{\mathrm{2}} \\ $$$${n}\left({n}+\mathrm{3}\right)=\mathrm{2}{m}^{\mathrm{2}} +\mathrm{1993} \\ $$$${LHS}\:{is}\:{always}\:{even},\:{even}×{odd}={even} \\ $$$${RHS}\:{is}\:{always}\:{odd},\:{even}+{odd}={odd} \\ $$$$\Rightarrow{no}\:{solution}\:{for}\:{n}! \\ $$
Commented by A.Haq.Soomro last updated on 15/Sep/18
V Nice MrWWW!
$$\mathrm{V}\:\mathrm{Nice}\:\mathrm{MrWWW}! \\ $$
Commented by MrW3 last updated on 15/Sep/18
thank you sir!  are you an old friend with a new ID?
$${thank}\:{you}\:{sir}! \\ $$$${are}\:{you}\:{an}\:{old}\:{friend}\:{with}\:{a}\:{new}\:{ID}? \\ $$

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