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Determine the integers $x$ and $y$, knowing that it verifies the relation:

$$x^3-x^2y+xy^2-y^3=2023$$

My idea

We can write the given relation as

$$(x^2+y^2)(x-y)=2023=7\cdot 17^2$$

I wonder if we can solve this problem in a different way than the method of testing for each possible option.

Hope one of you can help me! Thank you!

Lucky Chouhan
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IONELA BUCIU
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2 Answers2

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Hint

  • Each of $7, 7 \cdot 17, 7 \cdot 17^2$ is equivalent to $3$ modulo $4$ and so cannot be a sum of $2$ squares.
  • The Triangle Inequality gives $x - y \leq |x| + |y| \leq x^2 + y^2$.

These two restrictions leave a single possibility: $$x^2 + y^2 = 17^2, \qquad x - y = 7 .$$ N.b. there are two solutions $(x, y)$.

Travis Willse
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  • What does N.b. stand for? –  Oct 21 '23 at 17:00
  • @EtackSxchange https://en.wikipedia.org/wiki/Nota_bene – Travis Willse Oct 21 '23 at 17:19
  • @TravisWillse it tells no where that those numbers are the sidrs of a traingle. – IONELA BUCIU Oct 21 '23 at 17:41
  • @IonelaBuciu See https://en.wikipedia.org/wiki/Triangle_inequality#Example_norms – Travis Willse Oct 21 '23 at 18:05
  • @IonelaBuciu Even without knowing the triangle inequality you can see that $x^2+y^2=17^2$ is a Pythagorean triplet. Even if you aren't able to find $x$ & $y$ using this you can find $y$ from $x-y=7$ and substitute it into $x^2+y^2=289$ to find the integers. –  Oct 21 '23 at 20:22
  • Of course, one can separately check the three possibilities that are not eliminated by the first observations, but o.p. seemed to want to minimize the amount of checking individual cases. – Travis Willse Oct 21 '23 at 22:04
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Here is a way you can eliminate most of the options. I will write the possible pairs of values as $(a,b)$ with $x-y=a$ and $x^2+y^2=b$. With all possible values we have 6 options $$ (7, 17^2), (17^2, 7), (17, 7 \times 17), (7 \times 17, 17), (1, 2023), (2023, 1) $$

Now consider solving the equations generically. We first have from the first equation $$ x = y+a $$ Substituting into the second equation yields $$ (y+a)^2+y^2 = b $$ $$ 2y^2+2ay+(a^2-b)=0 $$ We know this will have solutions if the discriminant is greater than $0$, so we must have $$ (2a)^2-4(2)(a^2-b) \geq 0 $$ $$ a^2-2a^2+2b \geq 0 $$ which gives us $$ 2b \geq a^2$$

(Alternatively you can consider the equations in a geometric context and consider the closest point to the origin on the line to arrive at a similar result)

With this result we can eliminate 4 options leaving only $(7, 17^2), (1, 2023)$ as possible options. Then just solve each pair as simultaneous equations (note only one of the two gives integer solutions). Travis Willse answer provides a more efficient method but this is just another way of going about it.

Jacob Walker
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