a=\frac{62k+1}{k-1} or a = [62k+1] / [k-1] a, k are natural numbers I must find all a
Nov 1, 2011 #1 Patronas 3 0 [tex]a=\frac{62k+1}{k-1} [/tex] or a = [62k+1] / [k-1] a, k are natural numbers I must find all a Last edited by a moderator: Nov 6, 2011
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Nov 1, 2011 #2 dacruick 1,033 1 what do you mean you must find all a. Through what method must you find all a?
Nov 1, 2011 #3 mathman Science Advisor 7,898 461 For starters k must be even. If k is odd, k - 1 is even and 62k + 1 is odd.
Nov 1, 2011 #4 Patronas 3 0 I know that k must be even. But then, I calculate for k=2, k=4, k=6 .... And I have results a=63, 65, 69, 71, 83, 125 but i need good method
I know that k must be even. But then, I calculate for k=2, k=4, k=6 .... And I have results a=63, 65, 69, 71, 83, 125 but i need good method