a^3 + 2017a = b^3 -2017b diophantine

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a^3 + 2017a = b^3 -2017b diophantine

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Source: Mathematics Regional Olympiad of Mexico Northeast 2017 P3

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#1 h Sep 21, 2022, 7:18 PM • 1 Y

Y by Mango247

Prove that there is no pair of relatively prime positive integers $(a, b)$ that satisfy the equation
$a^3 + 2017a = b^3 -2017b.$

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#2 h Sep 21, 2022, 10:31 PM

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Note that $p\triangleq 2017$ is a prime. We rewrite the expression as
$p(a+b)=(b-a)(a^2+ab+b^2).$ Case 1. $a+b$ is odd. Note that $(a+b,b-a)=1$ . For this reason, $b-a\mid p$ , yielding $b-a\in\{1,p\}$ . If $b-a=p$ then $a+b=a^2+ab+b^2>a+b$ , a contradiction. Likewise if $b-a=1$ then $p(a+b) = (a^2+ab+b^2)\implies p(2a+1) = (3a^2+3a+1)$ . In particular, $2a+1\mid 3a^2+3a+1 \implies 2a+1\mid 6a^2+6a+2$ . Moreover, $2a+1\mid 6a^2+3a$ , yielding $2a+1\mid 3a+2\implies 2a+1\mid 6a+4$ . Likewise $2a+1\mid 6a+3$ , both of which collectively yield a contradiction.
Case 2. $a+b$ is even. In this case, $(a+b,b-a)=2$ . Hence, $(b-a)/2\mid p$ . Thus $b-a = 2$ or $2p$ . Both cases are handled analogously.

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