Find all n ∈ Z such that k = 1 + 4 n 5 , ( k ∈ Z )My question is rather general but I got stuck in that issue after trying to solve a trigonometric equation.After simplifying I got this: sin ⁡ ( 5 x 4 ) + cos ⁡ x = 2which is true if and only if { sin ⁡ ( 5 x 4 ) = 1 cos ⁡ x = 1 Hence cos ⁡ x = 1 ⟹ x = 2 π k , ( k ∈ Z ) and sin ⁡ ( 5 x 4 ) = 1 ⟹ 5 x 4 = π 2 + 2 π n = 2 π 5 + 8 π n 5 , ( n ∈ Z ) Thus the solutions are 2 π k = 2 π 5 + 8 π n 5 k = 1 + 4 n 5 , ( k ∈ Z ) Here I'm stuck. Clearly the constraint of k ∈ Z is not true for all n's. n has to be made up of some number m that makes the numerator divisible by 5. The answer that I'm given says that n = 5 m + 1, which makes sense but I don't know how to get there. And what if, just to say, k = 8 + 13 n 7 ?Thanks!!

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Find all   n  ∈      Z   such that   k  =            1      +      4      n        5    ,    (  k  ∈      Z    )My question is rather general but I got stuck in that issue after trying to solve a trigonometric equation.After simplifying I got this:  sin  ⁡      (                  5        x            4        )    +  cos  ⁡  x  =  2which is true if and only if                              {                                                    sin                ⁡                                  (                                                            5                      x                                        4                                    )                                                            =                1                                                                    cos                ⁡                x                                            =                1                                                                  Hence                    cos        ⁡        x                            =        1                                    ⟹                x                            =        2        π        k        ,                (        k        ∈                  Z                )            and                    sin        ⁡                  (                                    5              x                        4                    )                                    =        1                                    ⟹                                      5            x                    4                                    =                  π          2                +        2        π        n                                          =                              2            π                    5                +                              8            π            n                    5                ,                (        n        ∈                  Z                )            Thus the solutions are                    2        π        k                            =                              2            π                    5                +                              8            π            n                    5                                    k                            =                              1            +            4            n                    5                ,                (        k        ∈                  Z                )            Here I&#039;m stuck. Clearly the constraint of   k  ∈      Z   is not true for all n&#039;s. n has to be made up of some number m that makes the numerator divisible by 5. The answer that I&#039;m given says that   n  =  5  m  +  1, which makes sense but I don&#039;t know how to get there. And what if, just to say,       k    =                  8        +        13        n            7      ?Thanks!!

andytronicoh4t · Accepted Answer

If you&#039;re not familiar with congruences, you can look at it this way: If   4  n  +  1 is divisible by   5, then there is some integer r with   4  n  +  1  =  5  r, so that   4  n  =  5  r  −  1 and then   4  n  −  4  =  5  r  −  5. Factoring both sides gives   4  (  n  −  1  )  =  5  (  r  −  1  ). Since 4 and 5 are relatively prime, it follows that 5 divides   n  −  1; that is,   n  −  1  =  5  s for some integer   s and thus   n  =  1  +  5  s

Clara Dennis · Accepted Answer

k is an integer if and only if   n  =  5  m  +  1 for some   m  ∈      Z  To see this, solve the equation  4  n  +  1  ≡  0     (     m  o  d     5     )This is equivalent to  4  n  −  4  =  4  (  n  −  1  )  ≡  0     (     m  o  d     5  ), which immediately implies that   n  ≡  1     (     m  o  d     5     ) is necessary and sufficient.

Find all n in ZZ such that k = (1+4n)/(5), (k in ZZ)

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