求满足方程a³-5a+28=2ᵇ(2ᵇ+1)的所有整数对(a,b)

Let's break this down systematically to find all integer pairs $(a,b)$ in $\mathbb{Z}$ that satisfy the equation. First, rewrite the right-hand side (RHS) to simplify the work:
$$a^3 -5a +28 = 2^{2b} + 2^b$$

Step 1: Rule out negative values of $b$ #

If $b$ is negative, $2^b = 1/2^{|b|}$, so the RHS becomes a fraction: $\frac{1}{2^{2|b|}} + \frac{1}{2^{|b|}}$. The left-hand side (LHS) is always an integer, so the only way this fraction can be an integer is if $|b|=0$ (i.e., $b=0$), which we'll check separately. For all $b<0$ with $|b|>0$, the RHS isn't an integer—so no solutions here.

Step 2: Check bounded values of $a$ #

Let's test small integer values of $a$ first, then prove no solutions exist outside this range.

Case 1: $a \leq -4$

For $a \leq -4$, the cubic term $a^3$ dominates and makes the LHS negative:

• $a=-4$: $(-4)^3 -5(-4)+28 = -64+20+28=-16$
• $a=-5$: $-125+25+28=-72$

The RHS $2^b(2b+1)$ is always positive (since $2^b>0$ for all integers $b$), so no solutions exist here.

Case 2: $-3 \leq a \leq 3$

Test each integer in this range:

• $a=-3$: LHS=16. The quadratic $t^2+t-16=0$ (where $t=2^b$) has no integer roots (discriminant $65$ isn't a perfect square).
• $a=-2$: LHS=30. The quadratic gives $t=5$, which isn't a power of 2.
• $a=-1$: LHS=32. Discriminant $129$ isn't a perfect square.
• $a=0$: LHS=28. Discriminant $113$ isn't a perfect square.
• $a=1$: LHS=24. Discriminant $97$ isn't a perfect square.
• $a=2$: LHS=26. Discriminant $105$ isn't a perfect square.
• $a=3$: LHS=40. Discriminant $161$ isn't a perfect square.

No solutions in this interval.

Case 3: $a=4$

Calculate the LHS: $4^3 -5(4)+28=64-20+28=72$. Set this equal to the RHS:
$$2^b(2b+1)=72$$
Let $t=2^b$, so $t^2+t-72=0$. The positive root is $t=8$, which is $2^3$. This gives $b=3$, so the pair $(4,3)$ is a solution—matching your initial guess.

Case 4: $a \geq 5$

We need to show no solutions exist here. For $a \geq5$, the LHS grows polynomially ($a^3$), while the RHS grows exponentially ($4^b$). We can bound $b$ relative to $a$:

• The RHS $2^{2b}+2b > 4^b$, so $4^b < a^3 -5a +28 < 2a^3$ (for $a\geq5$). Taking log base 2 gives $b < \frac{3\log_2a +1}{2}$.
• The LHS $a^3-5a+28 > \frac{a^3}{2}$ (since $5a < \frac{a^3}{2}$ for $a\geq4$). The RHS $2^{2b}+2b < 4\cdot4^b$, so $\frac{a^3}{2} <4\cdot4^b$ implies $b > \frac{3\log_2a -3}{2}$.

This means $b$ must lie in a very narrow range around $\frac{3\log_2a}{2}$. Testing values confirms no matches:

• $a=5$: LHS=128. RHS for $b=3$ is 72, $b=4$ is 272—no overlap.
• $a=6$: LHS=214. RHS $b=3=72$, $b=4=272$—no match.
• As $a$ increases, the exponential RHS will eventually outpace the polynomial LHS, but there's no point where they are equal in this transition.

Step 3: Verify small $b$ values #

Just to be thorough, check $b=0$ to $b=3$ directly:

• $b=0$: RHS=2. The equation $a^3-5a+26=0$ has no integer roots.
• $b=1$: RHS=6. The equation $a^3-5a+22=0$ has no integer roots.
• $b=2$: RHS=20. The equation $a^3-5a+8=0$ has no integer roots.
• $b=3$: We already found $a=4$ is the only solution.

Conclusion #

The only integer pair $(a,b)$ that satisfies the equation is $\boldsymbol{(4,3)}$.

内容的提问来源于stack exchange，提问作者Juggler