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Source: HCM Specialized School’s Practical Math Entrance Exam (Round 2)

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#1 h Apr 28, 2025, 4:15 PM

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Let $a,b$ be integer numbers $\geq 3$ satisfy that: $a^2=b^3+ab$ . Prove that:
a) $a,b$ are even
b) $4b+1$ is a perfect square number
c) $a$ can’t be any power $\geq 3$ of a positive integer number

This post has been edited 2 times. Last edited by giangtruong13, May 2, 2025, 3:14 PM

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#2 h Apr 28, 2025, 5:08 PM • 1 Y

Y by giangtruong13

Part (c) must be a can't be a power $\ge 3$ as $(a,b)=(100,20)$ solves the system.

If $a$ is odd, $b^3+ab$ is even, a contradiction. So, both $a$ and $b$ are even. Let $d={\rm gcd}(a,b)$ with $a=da_1$ and $b=db_1$ and $(a_1,b_1)=1$ . We get $a_1^2=db_1^3+a_1b_1$ . In particular, $b_1\mid a_1^2$ so $b_1=1$ and $d=a_1^2-a_1$ . So, $4b+1=(2a_1-1)^2$ . Lastly, we show $a$ can't be a power. Suppose $a=u^\ell$ for $\ell\ge 2$ . Then, $a=da_1=a_1^2(a_1-1) = u^\ell$ . As $(a_1,a_1-1)=1$ , we must have that $a_1^2$ and $a_1-1$ are both powers, so $\ell=2t$ is even and $a_1 = x^t$ and $a_1-1=y^{2t}$ for some $x,y$ and $t\ge 2$ , impossible.

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#3 h Apr 28, 2025, 10:09 PM

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Part (a) three cases:
case 1 $a,b$ are odd:
$1^2 \equiv 1^3 + (1)(1) \pmod{2}$ $1 \equiv 0 \pmod{2}$ $.$
case 2 $a$ odd and $b$ even:
$1^2 \equiv 0^3 + (0)(1) \pmod{2}$ $1 \equiv 0 \pmod{2}$ case 3 $a$ even and $b$ odd:
$0^2 \equiv 1^3 + (0)(1) \pmod{2}$ $0 \equiv 1 \pmod{2}$ So we are done.

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#4 h Apr 29, 2025, 7:22 AM

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Here's part (b):
We have, $b^2(4b+1)=4b^3+b^2$ $(i)$
Again, we have: $a^2=b^3+ab$
So, $4a^2-4ab=4b^3$ $(ii)$
Substitute $(i)$ for $(ii)$ , we obtain: $b^2(4b+1)=4a^2-4ab+b^2=(2a-b)^2$ . And the rest is easy

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