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Fermat's Theorem won't help you. Or will it?

Let \(x\), \(y\) be coprime positive integers with \(xy > 1\), and let \(n\) be an even positive integer. Prove that \(x^n + y^n\) is not divisible by \(x + y\).

Note by Finn Hulse
2 years, 10 months ago

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Let \(x+y=t\)
Since \(x \equiv -y \mod t\), we have,
\(x^n+y^n \equiv 2y^n \mod t \)
If the above modulo equals \(0\), then \(t|2y^n\) which is clearly impossible since \(t\) does not divide \(2\) (\(xy>1 \implies t \ge 3\)) and \(t\) does not divide \(y^n\) (\(x\) and \(y\) are coprime). Ishan Singh · 2 years, 10 months ago

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@Ishan Singh Looks good, but you need to be slightly careful, especially if you haven't used all the conditions in the problem. (Given that it's an olympiad problem, and they tend to be provide just enough details.) If so, think about it. Is that a necessary condition? If yes, then you need to use it. If not, then you might want to explain why the condition isn't needed.

In this case, the statement is true for \( (x,y) = (1,1) \), which contradicts your claim that we can solutions "only if \( y = 0 \)". Calvin Lin Staff · 2 years, 10 months ago

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@Calvin Lin @Calvin Lin I guess this should be suffice. Ishan Singh · 2 years, 1 month ago

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Why would you need Fermat's Last Theorem? What is wrong with a 'remainder-factor-theorem' argument like @Ishan Singh 's? Am I missing something obvious? Mursalin Habib · 2 years, 10 months ago

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@Mursalin Habib I added that just to make you think a little bit. :D Finn Hulse · 2 years, 10 months ago

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I am pretty sure this is correct.

For any \(x^n + y^n\), where \(x\) and \(y\) are coprime positive integers and \(n\) is an even positive integer, the equation is irreducible, i.e. it can not be factorised. Hence it has no integer root. But since the question states that \(x\) and \(y\) are integers and the sum of two integers is always an integer, \(x^n + y^n\) is not divisible by \(x + y\). Sharky Kesa · 2 years, 10 months ago

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@Sharky Kesa Do not confuse a statement about divisibility of polynomials, with a statement about divisibility of integers. Calvin Lin Staff · 2 years, 10 months ago

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@Calvin Lin Yeah. Finn Hulse · 2 years, 10 months ago

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