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Show that the equation a^3 + (a+ 1)^3 + (a + 2)^3 + (a + 3)^3 + (a + 4)^3 + (a + 5)^3 + (a + 6)^3 = b^4 + (b + 1)^4

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in Olympiad by (76.5k points)

Show that the equation

a3 + (a+ 1)3 + (a + 2)3 + (a + 3)3 + (a + 4)3 + (a + 5)3 + (a + 6)3 = b4 + (b + 1)4

has no solutions in integers a, b.

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1 Answer

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a3 + (a + 1)3 + (a + 2)3 + (a + 3)3 + (a + 4)3 + (a + 5)3 + (a + 6)3 = b4 + (b + 1)4

Let a + 3 = x

so LHS = (x – 3)3 + (x + 3)3 + (x – 2)3 + (x + 2)3 + (x – 1)3 + (x + 1)3 + x3 

Now 7 divides LHS but

b4 + (b + 1)4 ≡ 1, 3, 6, 1, 6, 5, 1 (mod 7)

Hence the equation has no solutions in integers a & b.

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