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High School Math Grades 9-12, Ages 13-18

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High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

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#1 h Apr 2, 2025, 5:03 AM • 1 Y

Y by Chonkachu

Find all ordered pairs of nonzero integers $(a, b)$ that satisfy $(a^2+b)(a+b^2)=(a-b)^3.$

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MathPerson12321

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MathPerson12321

#2 h Apr 2, 2025, 5:12 AM

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Sol

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#3 h Apr 3, 2025, 12:21 AM

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MathPerson12321 wrote:

Sol

except you expanded (a-b)^3 wrong

a quick check of (1, -1) doesn’t work, for example

This post has been edited 1 time. Last edited by wuwang2002, Apr 3, 2025, 12:21 AM

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#4 h Apr 3, 2025, 12:50 AM

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wuwang2002 wrote:

MathPerson12321 wrote:

Sol

except you expanded (a-b)^3 wrong

a quick check of (1, -1) doesn’t work, for example

yeah it should be $a^{3}-3a^{2}b+3ab^{2}-b^{3}$

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#6 h Apr 4, 2025, 10:05 PM

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After expanding and moving everything over to the left, we can divide by $b$ .

$a^2b+a+2b^2+3a^2-3ab=0$

Treating the equation as a quadratic in $b$ gives:

$2b^2+a(a-3)b+a(3a+1)=0\implies b=\frac{-a(a-3)\pm\sqrt{a^2(a-3)^2-8a(3a+1)}}{4}=\frac{-a(a-3)\pm\sqrt{a^4-6a^3-15a^2-8a}}{4}$

Let $\Delta=a^4-6a^3-15a^2-8a$

$(a^2-3a-12)^2=a^4-6a^3-15a^2+72a+144\implies\Delta=(a^2-3a-12)^2-80a-144$ .

So if $a\geq-1$ then $\Delta<(a^2-3a-12)^2$ .
$(a^2-3a-13)^2=a^4-6a^3-17a^2+78a+169=\Delta-2a^2+86a+169<\Delta\implies 2a^2-86a-169>0\implies a>\frac{43+27\sqrt{3}}{2}\vee a<\frac{43-27\sqrt{3}}{2}$
So if $a\geq 45$ then $(a^2-3a-13)^2<\Delta<(a^2-3a-12)^2$ .

So if $a\leq-2$ then $\Delta>(a^2-3a-12)^2$ .
$(a^2-3a-11)^2=a^4-6a^3-13a^2+66a+121=\Delta+2a^2+74a+121>\Delta\implies 2a^2+74a+121>0\implies a>\frac{-37+7\sqrt{23}}{2}\vee a<\frac{-37-7\sqrt{23}}{2}$
So if $a\leq -36$ then $(a^2-3a-12)^2<\Delta<(a^2-3a-11)^2$ .

Now check in the range $-35\leq a\leq 44$ when $\Delta$ is square.
We get $a\in\{-1,0,8,9\}$ and so for $a,b\neq 0$ we have $(a,b)\in\{(-1,-1),(8,-10),(9,-6),(9,-21)\}$ .

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#7 h Apr 4, 2025, 11:07 PM

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how you so orz?

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#8 h Apr 5, 2025, 2:37 AM

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See USAMO 1987/1.

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