practice problems pending

Q1. Can product of eight consecutive integers be the perfect 4-th power of an integer?
S1.
n(n+1)(n+2)...(n+7)
rearrange:
n(n+7)(n+1)(n+6)(n+2)(n+5)(n+3)(n+4)
Let a = n^2 + 7n + 6
Then product becomes:
P = (a-6).a.(a+4).(a+6) = a^4 + 4a^3 -36a^2 - 144a
(a+1)^4 = a^4 + 4a^3 + 6a^2 + 4a + 1
This is clearly more than the above product. Why? Since a is positive.
(a+1)^4 > P
Can we show a^4 < P?
P = a^4 + 4a(a^2 - 9a - 36), a^2 - 9a - 36 = (a-12)(a+3) > 0  since a >= 14 for n = 1
=> a^4 < P < (a+1)^4 H.P. that it can't be 4th power.

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