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n-4-2n-3-2n-2-n-7-a-2-a-N-n-n-N-

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Question Number 199170 by tri26112004 last updated on 28/Oct/23
n^4 +2n^3 +2n^2 +n+7 = a^2 (a∈N) →n=¿ (n∈N)
n4+2n3+2n2+n+7=a2(a∈N)→n=¿(n∈N)
Commented by Rasheed.Sindhi last updated on 31/Oct/23
Great! ThanX Sir
Great!ThanXSir
Commented by Rasheed.Sindhi last updated on 30/Oct/23
•a^2 ≥(1)^4 +2(1)^3 +2(1)^2 +(1)+7 a^2 >13⇒a≥4^★ •n^3 +2n^2 +2n+1=((a^2 −7)/n)∈N •n^4 +2n^3 +2n^2 +n+7=a^2 ⇒2n^3 +2n^2 +n+7=a^2 −n^4 =(a−n^2 )(a+n^2 )∈N ⇒n^2 <a⇒n<(√a)
∙a2⩾(1)4+2(1)3+2(1)2+(1)+7a2>13⇒a⩾4★∙n3+2n2+2n+1=a2−7n∈N∙n4+2n3+2n2+n+7=a2⇒2n3+2n2+n+7=a2−n4=(a−n2)(a+n2)∈N⇒n2<a⇒n<a
Commented by Rasheed.Sindhi last updated on 30/Oct/23
^★ What′s a′s upper limit?
★What′sa′supperlimit?
Commented by Frix last updated on 30/Oct/23
SOLUTION Part 1 We can factorize it. n^4 +2n^3 +2n^3 +n−a^2 +7=0 n=t−(1/2) t^4 +(t^2 /2)−a^2 +((109)/(16))=0 It′s easy to get n=−(1/2)±((√(−1±2(√(4a^2 −27))))/2) [_(of the ± signs) ^(all 4 combinations) ] n∈R^+ ⇒ n=−(1/2)+((√(−1+2(√(4a^2 −27))))/2) ⇒ for a∈R 4a^2 −27≥0∧−1+2(√(4a^2 −27))≥0 ⇒ a≤−((√(109))/4)∨a≥((√(109))/4) a∈N ⇒ a≥3 No upper limit.
SOLUTIONPart1Wecanfactorizeit.n4+2n3+2n3+n−a2+7=0n=t−12t4+t22−a2+10916=0It′seasytogetn=−12±−1±24a2−272[ofthe±signsall4combinations]n∈R+⇒n=−12+−1+24a2−272⇒fora∈R4a2−27⩾0∧−1+24a2−27⩾0⇒a⩽−1094∨a⩾1094a∈N⇒a⩾3Noupperlimit.
Commented by Rasheed.Sindhi last updated on 30/Oct/23
Thanks Frix sir!
ThanksFrixsir!
Commented by Frix last updated on 30/Oct/23
SOLUTION Part 2 4a^2 −27=b^2 4a^2 −b^2 =27 (2a−b)(2a+b)=27 2a−b=1∧2a+b=27 ⇒ a=7∧b=13 ⇒ n=2 ★ 2a−b=3∧2a+b=9 ⇒ a=3∧b=3 ⇒ n∉N No other possibilities.
SOLUTIONPart24a2−27=b24a2−b2=27(2a−b)(2a+b)=272a−b=1∧2a+b=27⇒a=7∧b=13⇒n=2★2a−b=3∧2a+b=9⇒a=3∧b=3⇒n∉NNootherpossibilities.
Commented by Frix last updated on 30/Oct/23
I was in a hurry before, didn′t see this but it′s obvious...
Iwasinahurrybefore,didn′tseethisbutit′sobvious…
Commented by Frix last updated on 31/Oct/23
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Commented by tri26112004 last updated on 01/Nov/23
a≥4 and n<(√a) It′s not OK
a⩾4andn<aIt′snotOK
Answered by Frix last updated on 28/Oct/23
Just try a few values for n ⇒ n=2∧a=7
Justtryafewvaluesforn⇒n=2∧a=7
Commented by tri26112004 last updated on 29/Oct/23
Exactly, only n=2 satyfied
Exactly,onlyn=2satyfied
Answered by Rasheed.Sindhi last updated on 29/Oct/23
n^4 +2n^3 +2n^2 +n+7 = A^2 (A∈N) →n=¿ (n∈N) f(n)=n^4 +2n^3 +2n^2 +n+7 − A^2 =0 (A∈N) Let a,b,c,d are roots, at least one of which is natural number.say a∈N a+b+c+d=−2 ab+ac+ad+bc+bd+cd=2 abc+abd+acd+bcd=−1 abcd=7−A^2 ⇒a=((7−A^2 )/(bcd)) A=1,2,... A=1⇒a=(6/(bcd))⇒bcd=1,2,3,6⇒a=6,3,2,1 f(n) isn′t satisfied A=2⇒abcd=7−2^2 =3⇒a=(3/(bcd))⇒bcd=1,3⇒a=3,1 f(n) isn′t satisfied .... A=7⇒abcd=7−7^2 ⇒a=((−42)/(bcd)) ⇒bcd=−1,−2,−3,−6,−7,−14,−21,−42 ⇒a=42,21,14,7,6,3,2,1 (A,n)=(7,2) satisfy f(n) n=2: •2+b+c+d=−2 determinant (((b+c+d=−4))) •ab+ac+ad+bc+bd+cd=2 2(b+c+d)+bc+bd+cd=2 2(−4)+bc+bd+cd=2 determinant (((bc+bd+cd=10))) •abc+abd+acd+bcd=−1 2bc+2bd+2cd+bcd=−1 2(bc+bd+cd)+bcd=−1 2(10)+bcd=−1 determinant (((bcd=−21))) f(n)=(n−2)(n^3 +4n^2 +10n+21) =(n−2)(n+3)(n^2 +n+7) n=2 is only natural root. Not a good approach!
n4+2n3+2n2+n+7=A2(A∈N)→n=¿(n∈N)f(n)=n4+2n3+2n2+n+7−A2=0(A∈N)Leta,b,c,dareroots,atleastoneofwhichisnaturalnumber.saya∈Na+b+c+d=−2ab+ac+ad+bc+bd+cd=2abc+abd+acd+bcd=−1abcd=7−A2⇒a=7−A2bcdA=1,2,…A=1⇒a=6bcd⇒bcd=1,2,3,6⇒a=6,3,2,1f(n)isn′tsatisfiedA=2⇒abcd=7−22=3⇒a=3bcd⇒bcd=1,3⇒a=3,1f(n)isn′tsatisfied….A=7⇒abcd=7−72⇒a=−42bcd⇒bcd=−1,−2,−3,−6,−7,−14,−21,−42⇒a=42,21,14,7,6,3,2,1(A,n)=(7,2)satisfyf(n)n=2:∙2+b+c+d=−2b+c+d=−4∙ab+ac+ad+bc+bd+cd=22(b+c+d)+bc+bd+cd=22(−4)+bc+bd+cd=2bc+bd+cd=10∙abc+abd+acd+bcd=−12bc+2bd+2cd+bcd=−12(bc+bd+cd)+bcd=−12(10)+bcd=−1bcd=−21f(n)=(n−2)(n3+4n2+10n+21)=(n−2)(n+3)(n2+n+7)n=2isonlynaturalroot.Notagoodapproach!
Commented by Rasheed.Sindhi last updated on 29/Oct/23
Once we found n=2 ∧ A=7 We′ve to prove that there′s no other natural root of f(n). For this put A=7 in f(n) and then trying to solve f(n) whose one factor is already known: n−2 n^4 +2n^3 +2n^2 +n+7 − A^2 =0 (A∈N) n^4 +2n^3 +2n^2 +n+7 − 7^2 =0 n^4 +2n^3 +2n^2 +n−42=0 (n−2)(n+3)(n^2 +n+7)=0 ∴ n=2 is only natural root.
Oncewefoundn=2∧A=7We′vetoprovethatthere′snoothernaturalrootoff(n).ForthisputA=7inf(n)andthentryingtosolvef(n)whoseonefactorisalreadyknown:n−2n4+2n3+2n2+n+7−A2=0(A∈N)n4+2n3+2n2+n+7−72=0n4+2n3+2n2+n−42=0(n−2)(n+3)(n2+n+7)=0∴n=2isonlynaturalroot.

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