3x + 1 Problem: Iterative Operations Neither Cause Loops nor Diverge to Infinity

The 3x+1 problem is a problem of continuous iteration for integers. According to the basic theorem of arithmetic and the way of iteration, we derive a general formula for continuous iteration for odd integers. Through this formula, we can construct a loop iteration equation and obtain the result of the equation: the equation has only one positive integer solution. In addition, this general formula can be converted into a linear indeterminate equation. The process of solving this equation shows that the relationship between the iteration result and the odd number being iterated is linear. Extending this result to all positive even numbers, we get the answer to the 3x + 1 question.

On the exponential Diophantine equation mx +(m+1) y =(1+m+m 2) z

Let m > 1 be a positive integer. We show that the exponential Diophantine equation mx + (m + 1) y = (1 + m + m 2) z has only the positive integer solution (x, y, z) = (2, 1, 1) when m ≥ 2.

On the Diophantine equations z2 = k(k2 + 3) and z2 = k(k2 + 12)

This paper provides an analytical method of finding all the (positive, integral) solutions of the Diophantine equation z2 = k(k2+3). We also prove analytically that the Diophantine equation z2 = k(k2+12) has no positive, integer solution. J. Bangladesh Acad. Sci. 45(1); 127-129: June 2021

3x + 1 problem: Iterative operations neither cause loops nor diverge to infinity

The 3x+1 problem is a problem of continuous iteration for integers. According to the basic theorem of arithmetic and the way of iteration, we derive a general formula for continuous iteration for odd integers. Through this formula, we can construct a loop iteration equation and obtain the result of the equation: the equation has only one positive integer solution. In addition, this general formula can be converted into a linear indeterminate equation. The process of solving this equation shows that the relationship between the iteration result and the odd number being iterated is linear. Extending this result to all positive even numbers, we get the answer to the 3x + 1 question.

3x + 1 Problem: Iterative Operations Neither Cause Loops nor Diverge to Infinity

The 3x+1 problem is a problem of continuous iteration for integers. According to the basic theorem of arithmetic and the way of iteration, we derive a general formula for continuous iteration for odd integers. Through this formula, we can construct a loop iteration equation and obtain the result of the equation: the equation has only one positive integer solution. In addition, this general formula can be converted into a linear indeterminate equation. The process of solving this equation shows that the relationship between the iteration result and the odd number being iterated is linear. Extending this result to all positive even numbers, we get the answer to the 3x + 1 question.

On the solvability of the simultaneous Pell equations x2 − ay2 = 1 and y2 − bz2 = v12

Let [Formula: see text] be fixed positive integers such that [Formula: see text] is not a perfect square and [Formula: see text] is squarefree, and let [Formula: see text] denote the number of distinct prime divisors of [Formula: see text]. Let [Formula: see text] denote the least solution of Pell equation [Formula: see text]. Further, for any positive integer [Formula: see text], let [Formula: see text] and [Formula: see text], where [Formula: see text] and [Formula: see text]. In this paper, using the basic properties of Pell equations and some known results on binary quartic Diophantine equations, a necessary and sufficient condition for the system of equations [Formula: see text] and [Formula: see text] to have positive integer solutions [Formula: see text] is obtained. By this result, we prove that if [Formula: see text] has a positive integer solution [Formula: see text] for [Formula: see text] or [Formula: see text] according to [Formula: see text] or not, then [Formula: see text] and [Formula: see text], where [Formula: see text] is a positive integer, [Formula: see text] or [Formula: see text] and [Formula: see text] or [Formula: see text] according to [Formula: see text] or not, [Formula: see text] is the integer part of [Formula: see text], except for [Formula: see text]

A note on Jeśmanowicz’ conjecture for non-primitive Pythagorean triples

Let \((a, b, c)\) be a primitive Pythagorean triple parameterized as \(a=u^2-v^2, b=2uv, c=u^2+v^2\), where \(u>v>0\) are co-prime and not of the same parity. In 1956, L. Jesmanowicz conjectured that for any positive integer \(n\), the Diophantine equation \((an)^x+(bn)^y=(cn)^z\) has only the positive integer solution \((x,y,z)=(2,2,2)\). In this connection we call a positive integer solution \((x,y,z)\ne (2,2,2)\) with \(n>1\) exceptional. In 1999 M.-H. Le gave necessary conditions for the existence of exceptional solutions which were refined recently by H. Yang and R.-Q. Fu. In this paper we give a unified simple proof of the theorem of Le-Yang-Fu. Next we give necessary conditions for the existence of exceptional solutions in the case \(v=2,\ u\) is an odd prime. As an application we show the truth of the Jesmanowicz conjecture for all prime values \(u < 100\).

On the generalized Ramanujan-Nagell equation $ x^2+(2k-1)^y = k^z $ with $ k\equiv 3 $ (mod 4)

<abstract><p>Let $ k $ be a fixed positive integer with $ k &gt; 1 $. In 2014, N. Terai ^{[<xref ref-type="bibr" rid="b6">6</xref>]} conjectured that the equation $ x^2+(2k-1)^y = k^z $ has only the positive integer solution $ (x, y, z) = (k-1, 1, 2) $. This is still an unsolved problem as yet. For any positive integer $ n $, let $ Q(n) $ denote the squarefree part of $ n $. In this paper, using some elementary methods, we prove that if $ k\equiv 3 $ (mod 4) and $ Q(k-1)\ge 2.11 $ log $ k $, then the equation has only the positive integer solution $ (x, y, z) = (k-1, 1, 2) $. It can thus be seen that Terai's conjecture is true for almost all positive integers $ k $ with $ k\equiv 3 $(mod 4).</p></abstract>

A note on the exponential Diophantine equation (A^2n)^x+(B^2n)^y=((A^2+B^2)n)^z

Let A, B be positive integers such that min{A,B}>1, gcd(A,B) = 1 and 2|B. In this paper, using an upper bound for solutions of ternary purely exponential Diophantine equations due to R. Scott and R. Styer, we prove that, for any positive integer n, if A >B3/8, then the equation (A2 n)x + (B2 n)y = ((A2 + B2)n)z has no positive integer solutions (x,y,z) with x > z > y; if B>A3/6, then it has no solutions (x,y,z) with y>z>x. Thus, combining the above conclusion with some existing results, we can deduce that, for any positive integer n, if B ≡ 2 (mod 4) and A >B3/8, then this equation has only the positive integer solution (x,y,z)=(1,1,1).