dear-sir-W-Mjs-the-set-1-4-n-have-the-condition-that-if-two-different-elements-are-selected-and-2112-is-added-to-the-result-then-the-result-is-a-perfect-square-if-n-is-a-positif-number-then

FacebookTweetPin

Question Number 78399 by john santu last updated on 17/Jan/20

$$dearsirW,Mjs$$$$theset\{1,4,n\}havetheconditionthat$$$$iftwodifferentelementsare$$$$selectedand2112isaddedto$$$$theresult,thentheresult$$$$isaperfectsquareifnisa$$$$positifnumber.thenthenumber$$$$ofpossiblevaluesofnis$$$$\left(A\right)8\left(B\right)7\left(C\right)6\left(D\right)5$$$$\left(E\right)4$$

Commented by mr W last updated on 17/Jan/20

$$questionisnotclear!$$$$whatmeans“twodifferentelements$$$$areselectetand2112isaddedtothe$$$${result}^{″}whentwodifferentelements$$$$areselected,say1and4,whatisthe$$$$result?thesumofthesetwoelements$$$$ortheproductoftheseelementsor$$$$somethingelse?$$$$pleasemakethequestionclear.$$

Commented by john santu last updated on 17/Jan/20

$$thepurposeofthisproblem$$$$isif2membersareselected$$$$fromthesetforexample$$$$1andnthen1\times n+2112the$$$$resultarequadraticnumber$$

Commented by mr W last updated on 17/Jan/20

$$nowclear!sothequestionshouldbe$$$$\dots$$$$iftwodifferentelementsare$$$$selectedand2112isaddedto$$$$theirproduct,thentheresult$$$$\dots$$

Commented by john santu last updated on 17/Jan/20

$$sohowmuchisthevalueofn$$$$,sir?$$

Commented by john santu last updated on 17/Jan/20

$$igotinfinitysir.isright?$$

Commented by MJS last updated on 17/Jan/20

$$1\times 4+2112=2116={46}^2$$$$\Rightarrow n\mathrm{can}\mathrm{be}\mathrm{any}\mathrm{number}$$$$1\times n+2112=p^2\Rightarrow \mathrm{infinite}\mathrm{solutions}\mathrm{for}n$$$$4\times n+2112=p^2\Rightarrow \mathrm{infinite}\mathrm{solutions}\mathrm{for}n$$

Commented by john santu last updated on 17/Jan/20

$$yessir.igotthesameanswer$$

Commented by mr W last updated on 17/Jan/20

$$butwhatifnshouldalsobeaperfect$$$$square?$$$$1\times n+2112=p^2$$$$4\times n+2112=p^2$$

Commented by john santu last updated on 17/Jan/20

$$nispositivenumbersir$$

Commented by mr W last updated on 17/Jan/20

$$iknow.ifnisonlypositiveinteger,$$$$thereareinfinitepossibilities.but$$$$ifnshouldalsobeperfectsquare,$$$$thenithinkthepossibilitiesare$$$$finite!$$

Commented by MJS last updated on 17/Jan/20

$$\mathrm{let}n={(p-q)}^2$$$$\left(1\right){(p-q)}^2+2112=p^2\Rightarrow p=\frac{q^2+2112}{2q}$$$$p,q\in \mathbb{N}\wedge 0<q⩽p\Rightarrow 0<q⩽45$$$$\Rightarrow$$$$q\in \{2,4,6,8,12,16,22,24,32,44\}$$$$p\in \{529,266,179,136,94,74,59,56,49,46\}$$$$n\in \{277729,68644,29929,16384,6724,3364,1369,1024,289,4\}$$$$$$$$\left(2\right)4{(p-q)}^2+2112=p^2\Rightarrow q=p-\frac{\sqrt{p^2-2112}}{2}$$$$\Rightarrow$$$$p\in \{46,56,74,94,136,266\}$$$$q\in \{45,40,45,53,72,135\}$$$$n\in \{17161,4096,1681,841,256,1\}$$$$$$$$\mathrm{still}\mathrm{infinite}\mathrm{possibilities}\mathrm{for}n\mathrm{with}$$$$1\times 4+2112={46}^2$$

Commented by Rasheed.Sindhi last updated on 17/Jan/20

$$nshouldbesuchthat$$$$(1\times n+2112)\mathrm{\&}(4\times n+2112)$$$$\mathit{b}\mathit{o}\mathit{t}\mathit{h}areperfectsquares.$$

Commented by john santu last updated on 17/Jan/20

$$yessir.whatisvalueofnpossible?$$$$$$

Commented by john santu last updated on 17/Jan/20

$$inmybooktheansweris7$$

Answered by john santu last updated on 17/Jan/20

$$1\times n+2112=k^2\Rightarrow n=k^2-2112$$$$4\times n+2112=p^2\Rightarrow n=\frac{p^2-2112}{4}$$$$k^2-2112=\frac{p^2-2112}{4}$$$$k^2-\frac{p^2}{4}=2640\Rightarrow (k-\frac{p}{2})(k+\frac{p}{2})=2640$$

Commented by Rasheed.Sindhi last updated on 17/Jan/20

$$Butsirnispositive.$$

Commented by MJS last updated on 17/Jan/20

$$\mathrm{yes}.\mathrm{I}\mathrm{posted}all9\mathrm{solutions},\mathrm{then}\mathrm{I}\mathrm{wrote}\mathrm{the}$$$$\mathrm{last}2\mathrm{lines}\dots$$

Commented by MJS last updated on 17/Jan/20

$$\mathrm{error}\mathrm{in}\mathrm{last}\mathrm{line}$$$$k^2-\frac{p^2}{4}=1584$$$$\mathrm{your}\mathrm{idea}\mathrm{is}\mathrm{good},\mathrm{it}\mathrm{leads}\mathrm{to}$$$$n\in \{-512,-87,97,697,3072,8497,16113,37888,155497\}$$$$\mathrm{since}{\mathrm{we}}^{\prime }\mathrm{re}\mathrm{looking}\mathrm{for}n\in \mathbb{N}\mathrm{we}\mathrm{have}\mathrm{indeed}$$$$7\mathrm{solutions}$$

Commented by john santu last updated on 17/Jan/20

$$waw..thanksyousir$$

Answered by Rasheed.Sindhi last updated on 17/Jan/20

$$\Rightarrow \{\begin{array}{l}1\times 4+2112={46}^2\\ 1\times n+2112isperfectsquare\\ 4\times n+2112isperfectsquare\end{array}$$$$4\times n+2112isperfectsquare$$$$\Rightarrow n+528isperfectsquare$$$$\Rightarrow \{\begin{array}{l}n+2112isperfectsquare\\ n+528isperfectsquare\end{array}$$$$Letn+528=q^2$$$$n+2112=n+528+1584=p^2>q^2$$$$q^2+1584=p^2$$$$(p+q)(p-q)=1584=2^4.3^2.11$$$$p,q\in {\mathbb{Z}}^{+}\wedge p>q\Rightarrow p+q>p-q$$$$\bullet Forp,qbeingwholenumbers$$$$p+qandp-qarebotheven.$$$$$$$$p+q=44\Rightarrow p-q=36$$$$p+q=48\Rightarrow p-q=33\left(rejected\right)$$$$p+q=66\Rightarrow p-q=24$$$$p+q=88\Rightarrow p-q=18$$$$Continue$$

Commented by john santu last updated on 17/Jan/20

$$thanksyou$$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

FacebookTweetPin