💡 Problem Statement:
Find all pairs of integers \((a,b)\) such that: \(a^3 + b^3 = (a + b)^3 – 2020\)
✅ Step 1: Expand the Cubic Expression
First, expand the right-hand side of the equation: \((a + b)^3 = a^3 + b^3 + 3a^2b + 3ab^2\)
Now, substitute into the original equation: \(a^3 + b^3 = a^3 + b^3 + 3a^2b + 3ab^2 – 2020\)
Cancel the common terms on both sides: \(0 = 3a^2b + 3ab^2 – 2020\)
Factor the right-hand side: \((a+b)=2020\)
✅ Step 2: Check for Integer Solutions
Factor 2020:
\(2020 = 2^2 \times 5 \times 101\)
For the equation \(3ab(a+b)=2020\) to hold, the left-hand side must be divisible by \(3\). However, 2020 is not divisible by 3. This means that the equation cannot be satisfied with any integer values of \(a\) and \(b\)
✅ Step 3: Rearrange the Equation
Isolate the product term: \(ab(a + b) = \frac{2020}{3}\)
Since \(2020\) is not divisible by \(3\), this implies that there are no integer solutions under this form. Hence, we must reconsider the expansion.
✅ Step 4: Rearrange the Original Equation Differently
Rewrite the equation by expanding only part of the right-hand side: \(a^3 + b^3 = (a^3 + b^3) + 3ab(a + b) – 2020\)
Cancel the common terms again: \(0 = 3ab(a+b)−2020\)
Rearrange: \(3ab(a+b) = 2020\)
Thus: \(ab(a + b) = \frac{2020}{3}\)
Since \(2020\) is not divisible by \(3\), this indicates that no integer solutions exist. Therefore, we need to use numerical methods to test possible integer pairs.
✅ Step 5: Trial and Error with Suitable Integer Pairs
We now check integer pairs \((a,b)\) by substitution to see if they satisfy the original equation:
- For \((10,10)\)
\(10^3 + 10^3 = (10 + 10)^3 – 2020\)
\(1000+1000= 20^3 – 2020\)
\(2000 = 8000 – 2020\)
\(2000=5980(False)\) - For \((10,5)\) :
\(10^3 + 5^3 = (10 + 5)^3 – 2020\)
\(125 = 15^3 – 2020\)
\(125=3375−2020\)
\(125=1355(False)\)
- For \((10,−5)\):
\(10^3 + (-5)^3 = (10 – 5)^3 – 2020\)
\(1000 – 125 = 5^3 – 2020\)
\(875=125−2020\)
\(875=−1895(False)\)
- For \((20,−10)\):
\(20^3 + (-10)^3 = (20 – 10)^3 – 2020\)
\(8000−1000= 10^3 – 2020\)
\(7000=1000−2020\)
\(7000=−1020(False)\)
Conclusion
Since \(2020\) is not divisible by \(3\), there are no integer solutions to the given equation. Thus, the solution to the problem is:
\(\mathbf{No\ integer\ pairs\ satisfy\ the\ condition.}\)
Also Read About: IMO 1964 Math Problem: Finding Three-Digit Numbers Satisfying Two Conditions
Khairun
