A Tricky Diophantine Equation

Find the last three digits of the sum of all positive integers n2014n \leq 2014 for which there exist integers a,b,c,d,ea, b, c, d, e (not necessarily distinct) such that a2+b2+c2+d2+e2=(3n+1)2(232n+5).a^2+b^2+c^2+d^2+e^2= \left( 3^n + 1 \right) ^2 \left( 2 \cdot 3^{2n} + 5 \right) .

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