Quadratic diophantine equation

Question

I ran into this competition task while searching the internet, and I've been trying to tackle it for a while now.

The task is to solve the

14x^2+15y^2=7^2023 equation on the set of integer pairs.

I started like this:

14(x^2+y^2)+y^2=7*(7^2022)

I even figured out that x can be both even and odd, while y can only be odd.

I appreciate any help you can provide!

Answer 1

There are no integer solutions. The equation mod 3 is

2x² = 1

This is because 14 = 2 mod 3, 15 = 0 mod 3, and 7 = 1 mod 3. Next, multiply both sides by 2 to get x² = 2 mod 3. Finally, observe that 2 is not a square mod 3. Indeed, 1² and 2² are both 1 mod 3.

Note: A similar approach works reducing mod 5, where one again reaches x² = 2 and observes that 2 is not a square mod 5. For a deeper dive, look up the law of quadratic reciprocity.

Answer 2

14x² + 15y² = 7²⁰²³

The two terms 14x² and 7²⁰²³ are multiples of 7, so the remaining term 15y² must also be a multiple of 7.

Therefore y can be written y = 7y₁ for some integer y₁.

Thus our original equation becomes

(1) 14x² + 15(7y₁)² = 7²⁰²³

which we can rewrite as

14x² + 15·7²·(y₁)²= 7²·7²⁰²¹

The two terms 15·7²·(y₁)² and 7²·7²⁰²¹ are multiples of 7², so the remaining term 14x² must also be a multiple of 7².

Therefore x can be written x = 7x₁ for some integer x₁.

This our equation (1) becomes

(2) 14(7x₁)² + 15(7y₁)² = 7²⁰²³

which we can rewrite as follows (remembering that 14 = (2·7)):

(2·7)·7²·(x₁)² + 15·7²·(y₁)² = 7³ · 7²⁰²⁰

The two terms (2·7)·7²·(x₁)² and 7³ · 7²⁰²⁰ are multiples of 7³, so the remaining term 15·7²·(y₁)² must also be a multiple of 7³.

Therefore y₁ can be written y₁ = 7y₂ for some integer y₂....

Keep going until the 2023 counts down to zero! Good luck and thanks.