Show that 14x^2 + 15y^2 = 7^(2012) has no integer solutions. Hint: Show that if

Question

Show that 14x^2 + 15y^2 = 7^(2012) has no integer solutions. Hint: Show that if such a solution exists, then y is divisible by 7. Then, show that x is also divisible by 7. What happens to the equation after this?

Explanation / Answer

example Show that 14(x^2) + 15(y^2) = 7^1990 has no solutions in integers? answer Suppose that there were an integral solution. Without loss of generality, assume that x, y > 0. So, 15y^2 = 7^1990 - 14x^2 = 7(7^1989 - 2x^2). Therefore, 7 | 15y^2 ==> 7 | y^2 (since gcd(15, 7) = 1) ==> 7 | y, since 7 is prime. Writing y = 7k for some positive integer k yields 15(7k)^2 = 7^1990 - 14x^2 ==> 7 * 15k^2 = 7^1989 - 2x^2 ==> 2x^2 = 7(7^1988 - 15k^2) So, 7 | 2x^2 ==> 7 | x^2 (since gcd(2, 7) = 1) ==> 7 | x. Now, we can write x = 7j for some positive integer j and obtain 2(7j)^2 = 7(7^1988 - 15k^2) ==> 14j^2 + 15k^2 = 7^1988. Note that this is the same equation, but with the power of 7 diminished by 2. Clearly, we can repeat the above procedure many times until we obtain 14r^2 + 15s^2 = 1 for some positive integers r and s. However, this equation has no integer solutions: Clearly neither r nor s can equal 0; and otherwise 14r^2 + 15s^2 > 1. Hence, we have reached a contradiction. So, no integral (x, y) can exist for the given equation. I hope this helps!

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.