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Algebra

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17 Mar 2008 19:17:09 IST

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$\text {Let} \ a,b,c,d \in \mathbb{N} \\ \\ \text {such that} \ x^2-x(a^2+b^2+c^2+d^2+1) + (ab+bc+cd+da) = 0 \ \text{has an integer root} \\ \\ \text{Prove that the other root is also an integer, and that both roots are perfect squares}$

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Anand Hegde

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17 Mar 2008 19:30:27 IST

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there is no x term or is it a typing error?

Anand Hegde

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17 Mar 2008 19:32:56 IST

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typing error right?

Anand Hegde

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17 Mar 2008 19:37:53 IST

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$\alpha+\beta = a^2 + b^2 + c^2 + d^2 + 1$

$\beta = a^2 + b^2 + c^2 + d^2 + 1 - \alpha$

$\beta$ = integer - integer = integer

My solution looks very silly.......

ERAGON007

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17 Mar 2008 19:40:14 IST

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shouldn't it be (a²+b²+c²+d²+1)X

Hari Shankar

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17 Mar 2008 19:57:56 IST

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edited the qn. sorry

Anand Hegde

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17 Mar 2008 20:07:49 IST

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are the roots perfect squares? I checked for different values of a,b,c,d..not getting perfect squares.......maybe i have committed a calculation mistake.....will check again.

sandeep ramesh

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17 Mar 2008 20:53:35 IST

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question doesnt work for a,b,c,d = 1,2,3,4 resp

what is the correct question?

ERAGON007

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17 Mar 2008 20:57:36 IST

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it works for a=b=c=d =1 i think the que is now correct

sandeep ramesh

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17 Mar 2008 20:58:43 IST

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no man the question works for all a=b=c=d that can be easily proved but it doesnt work for many more like ive stated

gokul subramanian

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17 Mar 2008 20:59:54 IST

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i think it is obvious 1st part...sum of roots is an integer..1 root is integer so other root is integer!

Anand Hegde

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17 Mar 2008 21:02:24 IST

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bhatt sir......elliddeeri?

gokul subramanian

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17 Mar 2008 21:28:05 IST

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@anandghegde:
when u make corrections write "EDITED"

Anand Hegde

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17 Mar 2008 21:29:56 IST

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what have I edited?

Hari Shankar

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17 Mar 2008 21:30:23 IST

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See, a,b,c d should be chosen so that the roots are integers. Obviously this does not happen for every combination of a,b,c and d. No mistake in the qn.

sandeep ramesh

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17 Mar 2008 21:40:29 IST

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so must we prove that that happens only for a=b=c=d

I think only that will be true if the discriminant must be a perfect square

Hari Shankar

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17 Mar 2008 21:42:15 IST

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what about other configs of (a,b,c,d).

sandeep ramesh

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17 Mar 2008 21:42:51 IST

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cant see any other thing working

sandeep ramesh

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17 Mar 2008 21:44:29 IST

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i believe that it might come to the fact that they wont be integers if the confg isnt so

pardesi .svk

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18 Mar 2008 01:35:24 IST

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$\mathrm{We \ start \ with \ the \ obvious \ claim \ that \ for \ any \ naturals } \\ \alpha \mathrm{and} \beta \\ \mathrm{we \ have} (\alpha-1)(\beta-1) \geq 0 \Rightarrow \alpha \beta + 1 \geq \alpha + \beta \\ \mathrm{Here \ it \ is \ clear \ that \ since \ one \ root \ is \ an integer \ the\ other \ is \ also \ an \ integer. \ Let \ them \ be } \\ \alpha,\beta \\ \mathrm{then \ we \ have } \\ \alpha+\beta = \sum a^{2} +1, \alpha \beta =\sum ab \\ \mathrm{using \ the \ above \ inequality \ we \ get} \sum ab \geq \sum a^{2} \\ \mathrm{But \ from \ AM-GM \ it \ follows \ that} \sum a^{2} \geq \sum ab \Rightarrow \sum a^{2}=\sum ab \Rightarrow \sum \frac{(a-b)^{2}}{2} =0 \Rightarrow a=b=c=d \\ \mathrm{Thus \ we \ the \ given \ quadratic \ as } \\ x^{2}-(4a^{2}+1)x+4a^{2}=0 \Rightarrow (\alpha,\beta)=(1,4a^{2}) \mathrm{or} (4a^{2},1) \\ \mathrm{in \ either \ case \ we \ are \ done}$

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