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Solve for integers $x$, $y$ $$x^3= y^2-22$$ using only elementary techniques!

I managed to show that there are no solutions using $\sqrt{22}$, but I'd like to explain this to (trained) high school students.

A small note: subtracting $27$ seems to help, but not really.

EDIT: my reasoning in the quadratic field was wrong. Indeed, there is a solution $(3, 7) $ that one can deduce from the "subtract $27$" trick.

Andrea Marino
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  • It is a famous Mordell curve, and Keith Conrad has written a note how to solve this - see here. But you need some elementary number theory. – Dietrich Burde Oct 14 '21 at 09:29
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    Isn't $(x-3)(x^2+3x+9)=(y-7)(y+7)$ an interesting factorization ? –  Oct 14 '21 at 09:32
  • It is! When you have a $xy=zw$ type equation, its solutions are $x=AB, y=CD, z=AC, w=BD$ with $A, B, C, D$ pairwise coprimes. If you now impose the conditions resulting from your factorization - like $ w-z = 14$ - you will get a messy system that I couldn't solve by hand. That's where I got stuck. – Andrea Marino Oct 14 '21 at 10:34
  • Setting $a(x-3)=b(y-7)$ or $a(x-3)=b(y+7)$ reduces to quadratic equations. –  Oct 14 '21 at 10:41
  • Where do your a, b come from? However, that's a nice observation: if $y= x+4$ or $y=x-10$ you get a quadratic equation. – Andrea Marino Oct 14 '21 at 10:51
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    There's an obvious solution(s) when both sides of Yves' factorisation is zero. Then it remains to show there are no more. – Macavity Oct 14 '21 at 11:01
  • @DietrichBurde : it seems to me that the case $k=22$ is not addressed. However, it is very interesting to see such topic in a broader context: thank you for the reference! – Andrea Marino Oct 14 '21 at 11:08
  • Andrea, there are infinitely many $k$, so not every $k$ is explicitly mentioned, but you can see from the text, what to do with $k=22$; it works like other cases. – Dietrich Burde Oct 14 '21 at 11:10
  • Which case do you think can be extended to $k=22$? At a first sight, techniques are very specific to some $k$! – Andrea Marino Oct 15 '21 at 13:44

1 Answers1

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You may to substitute $27$ instead of subtracting it. $$x^3= y^2-22\implies y=\pm\sqrt{x^3+22}$$

By inspection, we can see that $\quad\sqrt{3^3+22}=\sqrt{27+22}=\sqrt{49}=\pm7$

suggesting one solution: $\quad (x,y)=(3,\pm7)$

For the $22$-surd alone, if $x=0$, then $y=\sqrt{22}$ which is not an integer. WolframAlpha shows that the only integer solution is $(3,\pm7)$.

poetasis
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  • add $8$ then $x^3+2^3 = y^2-4^2=(y-2)(y+2)$ or subtract $27$ then$x^3-3^3 = y^2-7^2=(y-7)(y+7)$ and you have some solutions at hand... – Gottfried Helms Oct 16 '21 at 15:53
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    @Gottfried Helms $\quad -22 + 8 \ne -4^2.\quad $ I do see now why you suggest subtracting $27$. Either of our ways require seeing number combos at a glance like reading words but I don't see yours as "jumping out" like my solution. It's a matter of perspective and mind tricks like when moon craters sometimes look like fuzzy mounds. – poetasis Oct 16 '21 at 16:32
  • upps, poetasis - that $4^2$ was a mindless shot... sorry. With the moon craters: that's a nice poetic last word, tonight, though the moon not visible... thanks for that! :-) – Gottfried Helms Oct 16 '21 at 18:03
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    @Gottfried Helms You mentioned “moon tonight” so I looked at your profile to confirm that you are many hours ahead of me in Germany. I am in Nebraska, USA. My specialty is Pythagorean triples and ways to find them. My “logo pic” is a composite of triples as shown here. Nice to meet a fellow math hobbyist. – poetasis Oct 16 '21 at 19:05
  • poetasis - a very nice puzzle that with the 6 triples; well it has already been solved :-) May we meet another time again at another question ... :-) – Gottfried Helms Oct 16 '21 at 19:32