Definitely not a complex!

Algebra Level 4

If k k is a positive integer, then k 7 7 + k 5 5 + 2 k 3 3 − k 105 \dfrac { { k }^{ 7 } }{ 7 } +\dfrac { { k }^{ 5 } }{ 5 } +\dfrac { { 2k }^{ 3 } }{ 3 } -\dfrac { k }{ 105 } is definitely a/an _________ \text{\_\_\_\_\_\_\_\_\_} .

Always 0 Integer Irrational Rational but not integer

Vishnu Kadiri by Vishnu Kadiri , 19, India

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2 solutions

Vishnu Kadiri
Sep 12, 2016

3 Helpful 0 Interesting 0 Brilliant 0 Confused

A simpler proof can be given by mathematical induction although I really appreciate your approach

Siddharth Yadav - 4 years, 9 months ago

Note that, as pointed out in other solution, we can write the expression as 15 k 7 + 21 k 5 + 70 k 3 − k 105 \frac{15k^7+21k^5+70k^3-k}{105} Now, from Fermat's little theorem, it follows that k 7 ≡ k ( m o d 7 ) , k 5 ≡ k ( m o d 5 ) , k 3 ≡ k ( m o d 3 ) k^7\equiv k (\mod 7),\ k^5\equiv k (\mod 5),\ k^3\equiv k (\mod 3) . Thus, 15 k 7 ≡ 15 k ( m o d 105 ) , 21 k 5 ≡ 21 k ( m o d 105 ) , 70 k 3 ≡ 70 k ( m o d 105 ) 15k^7\equiv 15k (\mod 105),\ 21k^5\equiv 21k (\mod 105),\ 70k^3\equiv 70k (\mod 105) . Thus, 15 k 7 + 21 k 5 + 70 k 3 − k ≡ 106 k − k ( m o d 105 ) ≡ 0 ( m o d 105 ) 15k^7+21k^5+70k^3-k\equiv 106k-k (\mod 105)\equiv 0\ (\mod 105) . Hence the expression given is an integer.

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