9 Detailed answers
1)Let be a positive integer. Find all positive integers such that .
Solution. otherwi . If , we should have , i.e., for some .
2) Let be two positive integers and let be an odd prime number such that
Prove that
Solution. By Fermat, , so
3) Show that the only positive integer value of for which is a perfect cube for all positive integers is
Solution.xmas是什么意思 If , is not a power of 迪马特奥(becau it is and either or modulo ).
Choo some odd prime . Now, take some with odd and notice that but can be anything we want modulo .
4) Let be an integer. Show that there exists infinitely many positive integers such that
Solution. If is not a power of , choo an odd prime and take . Then, for each not divisible by , we have . Also, if (and, thereby, ), then so the sum in question is divisible by .
If is a power of , then take an odd prime divisor of and repeat the above argument with instead of (the last term is, obviously, not a problem)
5) Show that has a prime divisor which isn't a divisor of
Solution. Fal as stated. , .
6) Let be a prime number, and and positive integers. Prove that if then
Solution. , so assume that is odd. Then , so it cannot be a square. But .
7) Find all positive integers for which there exist positive integers and such that and
Solution. must be odd since the sum of 2 squares is divisible by only if both squares are. If , then is odd and , which means that , so and . Now it is just cas.
a) . Then for all , and so
which gives an immediate contradiction.
b) . Then (or vice versa) and we get , meaning whence , so giving the only solution .
8) Let be positive integers such that is odd and is a prime. Prove that if then is a power of
Solution. , so . Now divide and by the highest power of they contain (it has to be the same). This may change and but not in our condition. Then u the LTE to get , so . If , we get , so and , which is less than for odd but the ca is impossible.
9) Let be a prime number. Solve the equation in the t of positive integers.
Solution. By Fermat, .holy grail
a) is odd. Then , so and, if , then . Thus, in this ca and , which is never an integer (becau it is strictly between and ).
b) . Then but, unless , either or is not a power of .
So, the only solution is .
10) Find all solutions of the equationin positive integers.
Solution. must be or prime (otherwi any nontrivial divisor of leather是什么意思will divide both and . Now, , is a solution, is not, so it suffices to consider the ca when is an odd prime.
If is odd, then if . is not a solution, so we can consider only even .
Then or . But the left hand side is at least (just count evens up to ), so , which is at least for . It remains to note that for .
Now comes the remaining finite trial and error part:
is good
is bad
gives which is far too large.
gives , which is too large too
gives some big as well.
11) For some positive integer the number is a perfect power of a prime. Prove that is a prime.
Solution. Assume , . Then with . Thus, by LTE, and but for , one has , which gives a contradiction.
12) Let be three positive integers with and witch是什么意思Show that if prime divisors of the numbers and be the same, then is a perfect power of
Solution. I failed to find a way to u the LTE here. The way I solved it is as follows. Let . Then and have the same prime divisors (this us the proof of the principle I mentioned rather than the statement itlf: when you repeat going from to , you stop when you get two equal numbers).
But then each prime dividing has to divide whence is a power of . If were even, the remainder of the LHS modulo would be 培训机构 英语, so is odd. Then and must be a power of 2 too.
13) Find the highest degree of for which divides the number
Solution. Using and , we get and . Since when , the answer is .
14) Let be a prime number and be a positive integer. Show that if for some positive integers we havethen
Solution. Since , we must have . Now, factoring out browrand writing , , we get .
tmbAssume that is odd. Take any prime divisor and let . If is odd, we get whence and , giving an immediate contradiction. If , we get , so customaryand , i.e., , which immediately implies .
If , we just notice that if , so is the only possibility unless , which is easy to outrule.
15) Find all positive integers such that where is a prime.
Solution. The ca is easily done ( has remainder , so is the only possibility), so assume that is odd. Then and, by LTE, . But unless , , which gives the cond solution.
16) Let be a square-free number. Show that there does not exist positive integers and such that
Solution. Fal as stated: for all large enough .
17) Let and be two positive real numbers such that for each positive integer the number is a positive integer. Show that and are both positive integers.
and 18) Let and be two positive rational numbers such that for infinitely many positive integers the number is a positive integer. Show that and are both positive integers.
Solution. They are very much alike, so I'll combine the solutions.
In 17, start with the obrvation that and are rational, so .
Now we are in the conditions of 18. Write where is the least common denominator of . If , take any prime divisor of . Then for infinitely many and cannot divide or (otherwi it would divide them both and we could reduce the fractions). Led be the least power such that . Then all tho 's are for some integer and we get (just not to consider parately). But grows only logarithmically in , so we get a contradiction.
19) Does there exist a positive integer such that has exactly prime divisors and divides
Solution. Note that by LTE. Thus, if has 1999 distinct prime divisors
, will work. Note that each divisor of is also a divisor of , so the t of prime divisors either grows without bound or saturates to some finite t . In the latter ca, we have where is the least integer such that . Thus, where is the product of all primes in , which is absurd.
20) Suppo that and are non-negative integers, and is a prime number. Prove that
wall street journal▪ ;
▪ is the smallest positive integer satisfying the congruence equation .
Solution. by LTE and .
Furthermore, but not , so if , we have (powers of have only divisors that are powers of 2 themlves). If , then , so .
21) Let be a prime. Find the maximum value of positive integer such that
Solution. Let . Then, since , , and , we get
22) Find all positive integers which are greater than and
Solution. Let be the least prime divisor of . Let be the least positive integer for which . Then and , so any prime divisor of divides and is less than . Thus, not to run into a contradiction, we must have . Now, if is odd, we have , so , which is impossible. Thus , is even, is odd and whence , which is possible only if , . Put with odd and rewrite the condition as . Let be the least prime divisor of (now, surely, odd). Let be the least positive integer such that . Then and
whence must be (or has a smaller prime divisor), so , which is impossible. Thus .
23) Let be distinct real numbers such that the numbers
Are all integers. Prove that are both integers.
Solution. If , then is either an integer or a half-integer, the latter ca being impossible becau then is not an integer. Otherwi, and are integers by problem 17, so is rational, so are rational but all rational square roots of integers are integer.
