IMO Shortlist NT

by KingSmasher3, Sep 6, 2013, 3:13 AM

1995 NT #1: Let $ k$ be a positive integer. Show that there are infinitely many perfect squares of the form $ n \cdot 2^k - 7$ where $ n$ is a positive integer.
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Clearly, for each $k,$ if there is one perfect square that is $-7\pmod{2^k}$ then there are infinitely many perfect squares. We proceed with induction on $k$. When $k=1,$ we have $1^2 \equiv -7\pmod{2^1}.$ Suppose that the result holds for some $k=j.$ Then we have $2^j | a^2+7$ for some $a.$ If $2^{j+1} \not{|} a^2+7,$ we try to find $b$ such that $2^{j+1} | (a+b)^2+7.$ It is clear that such a $b$ exists because we need for $2ab+b^2 \equiv 2^j \pmod{2^{j+1}},$ which has roots, such as $b=2^j.$ Our induction is complete. Therefore, the result holds for all $k.$

1999 NT #3: Prove that there exists two strictly increasing sequences $(a_{n})$ and $(b_{n})$ such that $a_{n}(a_{n}+1)$ divides $b^{2}_{n}+1$ for every natural n.
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Note that for each $a_n,$ if there exists a $b_n$ such that $a_{n}(a_{n}+1) | b^{2}_{n}+1,$ or equivalently, if $-1$ is a quadratic residue mod $a_{n}(a_{n}+1),$ then we can find a $b_n$ such that $b_n>b_{n-1}.$ Additionally, it is clear that if $-1$ is a quadratic residue of $a_n$ and $a_n+1$ then it is a quadratic residue of $a_n(a_n+1).$ It claim that this is true for all $a_n=5^{2k}, k\in \mathbb{Z}^+.$ In fact, we can generalize this to all primes $1\pmod{4},$ not just $5.$ Clearly, $-1$ is now a quadratic residue of $a_n+1.$ We now prove the stronger statement that $-1$ is a quadratic residue of all $a_n=5^k.$ We proceed by induction on $k.$ It is true for $k=1.$ Assume that it holds for some $k=j.$ It follows that $5^j | a^2+1$ for some $a.$ If $5^{j+1} \not{|} a^2+1,$ we find $b$ such that $5^{j+1} | (a+b)^2+1.$ Let $b=m5^j.$ Then we have $5^{j+1} | a^2+1+2ab+b^2,$ or $5 | (a^2+1)/5^j+2am.$ Since we can obviously find such an $m,$ our induction is complete, and we are done.

Main point(s): These problems demonstrate a trick in using induction to determine the existence of certain quadratic residues for powers.
This post has been edited 2 times. Last edited by KingSmasher3, Oct 10, 2013, 3:10 AM

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  • orz blog!!!

    by KevinChen_Yay, Dec 29, 2024, 1:39 AM

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    by Pluto1708, Mar 20, 2019, 2:25 PM

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    by Jiminhio 10, Jan 16, 2014, 2:11 PM

  • thanks haha

    by KingSmasher3, Sep 6, 2013, 3:15 AM

  • happy birthday

    by cire_il, Sep 3, 2013, 9:10 PM

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    by applepi2000, Jun 25, 2013, 12:31 AM

  • Hmm USAMO so hard

    by antimonyarsenide, Jun 25, 2013, 12:28 AM

  • dude you do problem 2s?
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    by applepi2000, Jun 25, 2013, 12:28 AM

  • Hmm USAMO so hard

    by antimonyarsenide, Jun 25, 2013, 12:28 AM

  • dude you do problem 2s?
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    i am not trolling

    by applepi2000, Jun 25, 2013, 12:27 AM

  • Hey look an excellent problem blog.

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    by yugrey, Apr 5, 2013, 1:55 AM

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    by meisepic, Mar 26, 2013, 6:47 AM

  • too much math

    by cire_il, Mar 26, 2013, 3:15 AM

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