Advanced P-6(BOMC-2)

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sourav das
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Advanced P-6(BOMC-2)

Unread post by sourav das » Sun Apr 01, 2012 12:22 am

Let $d$ be any positive integer not equal to $2, 5, $ or $13$. Show that one can find
distinct $a, b$ in the set {$2, 5, 13, d$} such that $ab − 1$ is not a perfect square.
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nafistiham
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Re: Advanced P-6(BOMC-2)

Unread post by nafistiham » Sun Apr 01, 2012 2:11 am

I am really confused. How can it be where $2\cdot5-1=3^2,2\cdot13-1=5^2,13\cdot5-1=8^2$
I think I am lost somewhere.please clarify.
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Niloy Da Fermat
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Re: Advanced P-6(BOMC-2)

Unread post by Niloy Da Fermat » Sun Apr 01, 2012 2:55 am

nafistiham wrote:I am really confused. How can it be where $2\cdot5-1=3^2,2\cdot13-1=5^2,13\cdot5-1=8^2$
I think I am lost somewhere.please clarify.
এই সেট থেকে ৬ টা $ ab-1 $ পাওয়া যাবে ।প্রমাণ করতে হবে এর ভিতর অন্তত ১টা (সবগুলো হওয়ার দরকার নেই) perfect square হবে না ।
Last edited by Niloy Da Fermat on Sun Apr 01, 2012 3:01 am, edited 1 time in total.
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Niloy Da Fermat
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Re: Advanced P-6(BOMC-2)

Unread post by Niloy Da Fermat » Sun Apr 01, 2012 2:59 am

sorry for using bangla.but without it, i possibly couldn't clarify.
an example:if $ 13d-1 $ is not perfect square, that is enough
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*Mahi*
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Re: Advanced P-6(BOMC-2)

Unread post by *Mahi* » Sun Apr 01, 2012 12:34 pm

General Hint:
Assume all of $2d-1,5d-1,13d-1$ are perfect squares and then try finding a contradiction.
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