Abstract We consider the Diophantine equation x2-kxy+ky2+ ly = 0 for l = 2n and determine for which values of the odd integer k, it has a positive integer solution x and y.

Let l be a given nonzero integer. The authors give an explicit characterization of the positive integer k that makes the Diophantine equation x2 − kxy + y2 +lx = 0 have infinitely many positive integer solutions (x, y).

On the Diophantine Equation x2 – kxy + ky2 + ly = 0, l = 2n

On the Diophantine Equation x2-kxy+y2+lx=0.

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In this study, we investigate positive integer solutions of the Diophantine equations x^[email protected]?y^[email protected]?x=0 and x^2-kxy-y^[email protected]?y=0. It is shown that when k>3, x^2-kxy+y^2+x=0 has no positive integer solutions but the equation x^2-kxy+y^2-x=0 has positive integer solutions. Moreover, it is shown that the equations x^2-kxy-y^[email protected]?x=0 and x^2-kxy-y^[email protected]?y=0 have positive solutions when k>=1.

background

Solutions of some quadratic Diophantine equations

We prove that the equation x2 − kxy + y2 + x = 0 with k ϵ N+ has an infinite number of positive integer solutions x and y if and only if k = 3. For k = 3 the quotient x y is asymptotically equal to (3 + √5)/2 or (3 − √5)/2. Results of the paper are based on data obtained via Computer Algebra System ( derive 5). Some derive procedures presented in the paper made it possible to discover interesting regularities concerning simple continued fractions of certain numbers.

Infinitely many positive solutions of the diophantine equation x2 − kxy + y2 + x = 0

Let l be a given nonzero integer. The authors give an explicit characterization of the positive integer k that makes the Diophantine equation x
 2 − kxy + y
 2 +lx = 0 have infinitely many positive integer solutions (x, y).

On the Diophantine Equation x2 – kxy + ky2 + ly = 0, l = 2n

Citations

On the Diophantine Equation x2-kxy+y2+lx=0.

References

Solutions of some quadratic Diophantine equations

Infinitely many positive solutions of the diophantine equation x2 − kxy + y2 + x = 0

On the Diophantine equation x 2 − kxy + y 2 + lx = 0

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