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 \(x + y\).

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TopNewestLet \(x+y=t\)

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 · 3 years ago

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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 · 3 years ago

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@Calvin Lin I guess this should be suffice. – Ishan Singh · 2 years, 3 months 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 · 3 years ago

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– Finn Hulse · 3 years ago

I added that just to make you think a little bit. :DLog in to reply

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

alwaysan integer, \(x^n + y^n\) is not divisible by \(x + y\). – Sharky Kesa · 3 years agoLog in to reply

– Calvin Lin Staff · 3 years ago

Do not confuse a statement about divisibility of polynomials, with a statement about divisibility of integers.Log in to reply

– Finn Hulse · 3 years ago

Yeah.Log in to reply