Iran-NMO-2010-1
- Tahmid Hasan
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Let $a,b$ be two positive integers and $a>b$.We know that $\gcd (a-b,ab+1)=1$ and $\gcd (a+b,ab-1)=1$. Prove that $(a-b)^2+(ab+1)^2$ is not a perfect square.
বড় ভালবাসি তোমায়,মা
Re: Iran-NMO-2010-1
It is easy to show that $(a-b)^2+(ab+1)^2=(a^2+1)(b^2+1)$. If we can prove that $(a^2+1,b^2+1)=1$ then we're done.
Let $d=(a^2+1,b^2+1)$. Assume to the contrary that $d\neq 1$. Now $d|(a^2+1)-(b^2+1)=(a+b)(a-b)$. As $d>1$, let $p$ be a prime divisor of $d$. Now $p$ divides either $(a+b)$ or $(a-b)$. If $p|(a-b)$ then since $p|(a^2+1)(b^2+1)\Rightarrow p|(ab+1)\Rightarrow p|(a-b,ab+1)=1$, a contradiction. Similar contradiction can be showed if we assume that $p|(a+b)$. So we were wrong assuming that $d\neq 1$. So $d=1$ and that implies that both $a^2+1$ and $b^2+1$ are perfect squares which is impossible.
Let $d=(a^2+1,b^2+1)$. Assume to the contrary that $d\neq 1$. Now $d|(a^2+1)-(b^2+1)=(a+b)(a-b)$. As $d>1$, let $p$ be a prime divisor of $d$. Now $p$ divides either $(a+b)$ or $(a-b)$. If $p|(a-b)$ then since $p|(a^2+1)(b^2+1)\Rightarrow p|(ab+1)\Rightarrow p|(a-b,ab+1)=1$, a contradiction. Similar contradiction can be showed if we assume that $p|(a+b)$. So we were wrong assuming that $d\neq 1$. So $d=1$ and that implies that both $a^2+1$ and $b^2+1$ are perfect squares which is impossible.
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- Nadim Ul Abrar
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Re: Iran-NMO-2010-1
পীথাগোরিয়ান ট্রিপল এর ক্ষেত্রে , Primitive ট্রিপল থাকে একটা । বাট ,
$(a-b)^2+(ab+1)^2=(a+b)^2+(ab-1)^2$
$(a-b)^2+(ab+1)^2=(a+b)^2+(ab-1)^2$
$\frac{1}{0}$
- Tahmid Hasan
- Posts:665
- Joined:Thu Dec 09, 2010 5:34 pm
- Location:Khulna,Bangladesh.
Re: Iran-NMO-2010-1
নাদিম ভাই দেখি আমার মতই ওয়ান লাইনার মেরে দিলেন। আওপ্সে এই প্রব্লেমটাতে মাসুম ভাইও একই সমাধান দিছেNadim Ul Abrar wrote:পীথাগোরিয়ান ট্রিপল এর ক্ষেত্রে , Primitive ট্রিপল থাকে একটা । বাট ,
$(a-b)^2+(ab+1)^2=(a+b)^2+(ab-1)^2$
বড় ভালবাসি তোমায়,মা
Re: Iran-NMO-2010-1
এইটার প্রুফ দিতে হবে না? :/Nadim Ul Abrar wrote:পীথাগোরিয়ান ট্রিপল এর ক্ষেত্রে , Primitive ট্রিপল থাকে একটা ।
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Nur Muhammad Shafiullah | Mahi
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- Nadim Ul Abrar
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Re: Iran-NMO-2010-1
কনটেস্টে অবশ্যই দেয়া উচিত, আমার মনে হয় না এইটা খুব বেশি ট্রিভিয়াল কিছু।Nadim Ul Abrar wrote:আচ্ছা এইসব জিনিস কি প্রফ কইরা ইউস করা লাগে ??
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- Phlembac Adib Hasan
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Re: Iran-NMO-2010-1
@ Nadim ভাই, পিথাগোরিয়ান ট্রিপল দিয়ে আমিও করেছি তবে অন্যভাবে। মানে $a-b=2mn, ab+1=m^2-n^2$ এভাবে আরকি। কিন্তু আপনার এক লাইন থেকে সল্যুশন ঠিক বুঝলাম না। এরকমটা তো হতেই পারে, যেমন-
$(19^2-4^2)^2+(2\times 19\times 4)^2=377^2=(16^2-11^2)^2+(2\times 16\times 11)^2$
$(19^2-4^2)^2+(2\times 19\times 4)^2=377^2=(16^2-11^2)^2+(2\times 16\times 11)^2$
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- Nadim Ul Abrar
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- Phlembac Adib Hasan
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Re: Iran-NMO-2010-1
My solution:
From the identity of Nadim vai, we can find two different Pythagorean triples.
Case 1: $a-b$ even
So $a-b=2mn;ab+1=m^2-n^2$
Since $a-b$ and $a+b$ have same parity, $a+b=2xy;ab-1=x^2-y^2$
and $m^2+n^2=x^2+y^2\Longrightarrow m^2-x^2=y^2-n^2$
Therefore, $m^2-n^2=x^2-y^2+2\Longrightarrow 2(m^2-x^2)=2\Longrightarrow m^2-x^2=1$
which is strictly impossible.
Case 2: $a-b$ odd
$a-b=m^2-n^2;ab+1=2mn$
$a+b=x^2-y^2;ab-1=2xy$
and $m^2+n^2=x^2+y^2$
We get $2mn-2xy=2\Longrightarrow mn-xy=1$
Therefore, $(m+n)^2-(x+y)^2=2$
still impossible like the previous and we're done.
From the identity of Nadim vai, we can find two different Pythagorean triples.
Case 1: $a-b$ even
So $a-b=2mn;ab+1=m^2-n^2$
Since $a-b$ and $a+b$ have same parity, $a+b=2xy;ab-1=x^2-y^2$
and $m^2+n^2=x^2+y^2\Longrightarrow m^2-x^2=y^2-n^2$
Therefore, $m^2-n^2=x^2-y^2+2\Longrightarrow 2(m^2-x^2)=2\Longrightarrow m^2-x^2=1$
which is strictly impossible.
Case 2: $a-b$ odd
$a-b=m^2-n^2;ab+1=2mn$
$a+b=x^2-y^2;ab-1=2xy$
and $m^2+n^2=x^2+y^2$
We get $2mn-2xy=2\Longrightarrow mn-xy=1$
Therefore, $(m+n)^2-(x+y)^2=2$
still impossible like the previous and we're done.
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