# 250 problems in elementary number theory

250 Problems in Elementary Number Theory

Hogatt that every positive integer is a sum of distinct terms of Fibonacci sequence. Prove that every integer can be represented as a sum of five cubes of integers in infinitely many ways. Prove that the number 3 can be represented as a sum of four cubes of integers different from and 1 in infinitely many ways.

Prove by elementary means that there exist infinitely many positive integers which can be represented as sums of four squares of different in- tegers in at least two ways, and that there exist infinitely many positive in- tegers which can be represented in at least two ways as sums of four cubes of different positive integers. Prove that for positive integers m, in each representation of the number 4 m 7 as a sum of four squares of integers ;; 0, each of these numbers is ;; 2 m - I Prove that there exist infinitely many positive integers which cannot be represented as sums of two cubes of integers, but can be represented as sums of two cubes of positive rational numbers.

Prove that there exist infinitely many positive integers which can be represented as differences of two cubes of positive integers, but cannot be represented as sums of such two cubes. LlMS 21 positive integers which can be represented as differences of two kth powers of positive integers, but cannot be represented as sums of two kth powers of positive integers.

Find the number of decimal digits of the number -1 this is the largest known perfect number. Prove that the number 3!!! Prove the theorem of P. Erdos and M. Therefore, it is necessary and sufficient for t to be an integer divisor of 24, hence t must be equal to one of the numbers I, 2, 3, 4, 6, 8, 12, We shall prove the assertion by induction. The proof by induction follows immediately. Obviously, it suffices to show that each of the primes 11, 31, and 61 divides 20 15 Thus 15 Thus, 15 It follows that d is the greatest common divisor of a-I and m, which was to be proved. It follows from these formulas that which proves the desired property. On the other hand, if n is even, let 2 S be the highest power of 2 which divides n thus, s is a positive integer.

Thus one and only one of the numbers an and b n is divisible by 5. Then each of the numbers x, XX, x",x, In fact, for even x, each of the terms of the sequence x, x" , xr, Thus, n'1.! It follows that where t is a positive integer. We shall prove first the following theorem due to O. This proves the truth of the lemma. However, C. See a solution to my problem in Elemente der Mathematik, 18 , p. It suffices therefore to consider only numbers r equal to 0, 1, 2, By Euler's theorem, we have then nI29'C n We shall prove it by induction.

Thus we have nI9"-I. Clearly, n cannot be divisible by 3. The numbers of this form are called Cullen numbers. It is not known whether there exist infinitely many prime Cullen numbers. For positive integers n we have obviously qJ n!

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Thus ql-1 q Therefore 4. We find easily that the remainders upon dividing by 65 are I, 5, 13,29,61,60, 58, 54,46,30,63,64, and none of these remainders is zero. It is known see, for instance, Sierpiilski, [37, p. Thus s61 This assertion would follow from a conjecture of A. Schinzel concerning prime numbers . For prime num- bers n, both relations nI2"-2 and nI3"-3 hold because of Fermat's theorem. In the solution to Problem 27 we proved that ,t3 Rotkiewicz proved that there exist infinitely many positive integers n, both even and odd, such that nI2"-2 and n,t 3" Cipolla proved that for every positive integer a there exist infinitely many composite numbers n such that nia"-a.

See . From a certain conjecture of A.. Schinzel concerning prime numbers  it follows that there are infinitely many such composite numbers. The cube of an integer which is not divisible by 3 gives remainder 1 or -1 upon dividing by 9. It is easy to see that the square of an integer not divisible by 7 gives, upon dividing by 7, the remainders 1, 2 or 4.

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The square of an integer not divisible by 7 gives upon dividing by 7 the remainder 1, 2, or 4, hence the sum of such squares gives the remainder 1, 2, 3, 4, 5, or 6. The number y is not divisible by x since y is odd, while x is even. In fact, two last digits of every number divisible by must be zero, and the sum of digits of every number smaller than is obviously smaller than See Kaprekar . On the other hand, it is clear that the sum of decimal digits of the number n equals s.

By assumption concern- ing s we have g ;; h. The theorem being true for every number mq, we obtp.. Let'l, '2, and'3 be the remainders upon dividing the integers -a, -b, and -c by n. Therefore F" 'J! Banachiewicz suspected that this relation led P. Let a ,and b be two different integers.

For k sufficiently large, n will be positive integer. In case h :F 1, let Ql, Now, Jet i denote one of the numbers 1, 2, Since there are infinitely many such numbers n, the proof is complete. We shall present the proof based on an idea of A. Schinzel see . We do not know whether every even number is a difference of two primes. From a certain conjecture on prime numbers of A. Schinzel  , it follows that every even number can be represented as a difference of two l'rimes in infinitely many ways. We shall present the proof given by A. If u" is the 'nth term of the Fibonacci sequence.

The least such nunlber n is equal to 6. The numbers m!

This, in view of dlm! Sierpinski [37, p. Thus, all rectangular triangles whose sides are integers forming an arithmetic progression are obtained by increasing integer number of times the triangle with sides 3, 4, 5.

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Reutter, Elemente der Math. See . For prime num- bers n, both relations nI2"-2 and nI3"-3 hold because of Fermat's theorem. Sort order. As it was checked by A. Elementary number theory by David M. Since you apply not connected books, Pages, or rural mids, you may find from a creative pain.

Thus both progressions with difference 2 contain infinitely many triangular numbers. It is necessary and sufficient for b to be a quadratic residue for modulus a. We shall give the proof due to A. Let Pk denote the kth successive prime. Catalan's problem whether there are any other pairs of such integers is open.

The solution follows immediately from Problem 58, it can, however, be proved in a simpler way. Therefore our progression contains infinitely many terms relatively prime with m, which was to be proved. Let b be the first term of our progression and let a be its difference; thus the numbers a and b are positive integers. On the other hand, UII! It follows that there are no four terms of the Fibonacci sequence which form an arithmetic progression. We know [27, p. Since for every positive integer m 7 all possible remainders modulo m appear in the above sequences, we see that each of the arithmetic progressions with the difference m 7 contains infinitely many terms of the Fibonacci sequence.

It follows that and 8IuI4-u2. Suppose now that these two formulas hold for some positive integers n.

From the first fourteen remainders modulo 8 we see that these remainders may be only 0, 1, 2, 3, 5, and 7. These are the progressions of integer terms with the desired property, and with the least possible difference. It follows that the sequence of remainders modulo 11 of the Fibonacci sequence is periodic, with period 10; we easily find this sequence to be 1,1,2,3,5,8,2,10,1,0, The number 4 and also 6, 7, and 9 does not appear in this sequence.

The numbers are therefore pairwise relatively prime. Thus we defined by induction the infinite sequence kb k 2 , Thus, in our progression there exist infinitely many terms with the same set of prime divisors, which was to be proved see P6lya . Suppose now that the theorem is true for some positive integer s. By induction it follows that the theorem is true for every positive integer s, which was to be proved.

Therefore there exists only one arithmetic progression of difference 10 consisting of three primes, namely 3, 13, From a certain conjecture of A. On the other hand, from a conjecture of A.

If the difference of our progression were odd, then every second term of our progression would be even, which is impossible if our progression is to be formed of ten primes. Thus, the difference must be even.