I have the code for a loop to run. Right now it can check if a number is a square and at the same time check if it is congruent 0 mod 47. My questions is, how can I alter the code to see if it is congruent for more than one prime. SO for instance I would like it to check that a number is congruent, 0 mod 5, 0 mod 13 and 0 mod 17 all at once and only keep or print the numbers which are. I would try something like: 'Select[Range[10^7]^2, Divisible[7 # + 4, 5], Divisible[7 # + 4, 13], Divisible[7 # + 4, 17] &]' but it just prints numbers most of which I am assuming do not satisfy the desired effect
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2 Answers
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5 To find solutions, for example, the code:
FindInstance[{n == a^2, 7 n + 4 == m, m == 5 b}, {n, m, a, b}, Integers] returns {} to indicate no solutions. The alternative code
Reduce[{n == a^2, 7 n + 4 == m, m == 5 b}, {m}, Integers] returns False similarly. Try variations of these codes depending on your needs.
However, both Reduce[] and FindInstance[] are not good with exponential equations.
For that, you can use some code
Select[2^Range[0, 20] - 1, Divisible[#, 5] && Divisible[#, 13] &] which returns {0, 4095} and you can try variations of this.
- $\begingroup$ Thank you, I just want to adjust it so that it does it for numbers of the form (4^n)-1 so I tried the following code and it does nothing: $\endgroup$argamon– argamon2019-03-20 23:29:14 +00:00Commented Mar 20, 2019 at 23:29
- $\begingroup$ Select[Table[Divisible[(4^i)-1,5] && Divisible [(4^i)-1, 13] && Divisible[(4^i)-1,17] {i,200} &] $\endgroup$argamon– argamon2019-03-20 23:29:21 +00:00Commented Mar 20, 2019 at 23:29
- $\begingroup$ That is helpful but how could I use that to check for it being divisible by 5 and 13 and 17 at the same time or if I change the base 4 to 3 lets say? $\endgroup$argamon– argamon2019-03-20 23:49:22 +00:00Commented Mar 20, 2019 at 23:49
- $\begingroup$ I tried adjusting the base and it tells me that everything is zero: $\endgroup$argamon– argamon2019-03-20 23:51:16 +00:00Commented Mar 20, 2019 at 23:51
- $\begingroup$ {In[3]:= FindInstance[{2^i-1 == n, n == 5 a}, {i, a, n}, Integers] Out[3]= {{i -> 0, a -> 0, n -> 0}} $\endgroup$argamon– argamon2019-03-20 23:51:24 +00:00Commented Mar 20, 2019 at 23:51
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For the question, expressed in a comment by argamon, of finding numbers of the form (4^n)-1 congruent to 0 mod 5, 0 mod 13, and 0 mod 17, please consider the function ChineseRemainder.
Block[{c}, Flatten[Table[ c = ChineseRemainder[{0, 0, 0}, {5, 13, 17}, m]; If[c == m, m, {}], {m, Table[4^k - 1, {k, 1, 50}]}]] ] {16777215, 281474976710655, 4722366482869645213695, 79228162514264337593543950335}
Select[Range[10^5]^2, Divisible[#, 5] && Divisible[#, 13] && Divisible[#, 17] &]? $\endgroup$Select. The only thing you did wrong is not to useAnd(&&). The selection function is required to produce eitherTrueorFalse. And, of course, you may use any list you like as first argument ofSelect. $\endgroup$