Problem Statement:
Find all pairs of integers \((x,y)\) such that: \(x^2 + y^2 + 7 = 3xy\)
🧠 Step 1: Rearrange the Given Equation
First, rearrange the given equation: \( x^2 + y^2 + 7 = 3xy\)
Bring all terms to one side: \( x^2 + y^2 – 3xy + 7 = 0\)
🔍 Step 2: Use Completing the Square Technique
Group the variables \(x\) and \(y\)together:
\(x^2 – 3xy + y^2 = -7\)
We aim to complete the square. Recall the square expansion:
\((x – y)^2 = x^2 – 2xy + y^2\)
We need \(−3xy\), not \(−2xy\),so we split the middle term:
\( x^2 – 3xy + y^2 = (x – y)^2 – xy\)
Thus, the equation becomes:
\((x – y)^2 – xy = -7\)
Rearrange it:
\((x – y)^2 = xy – 7\)
🔢 Step 3: Try Specific Integer Values
Let us check for integer pairs (x,y) that satisfy the condition.
- For x=1:
\((1 – y)^2 = 1y – 7\)
\((1−y)^2=y−7\)
Try y = 2 :
\((1 – 2)^2 = 2 – 7\)
\(1 = −5 (False)\)
- For x = 2 :
\((2 – y)^2 = 2y – 7\)
Try y = −2 :
\((1 – (-2))^2 = -2 – 7\)
\(3^2 = -9 (False)\)
Try y = 2 :
\((2 – 2)^2 = 2(2) – 7\)
\(0 = 4−7\)
\( 0 = -3(False)\)
Try y = 3 :
\((2 – 3)^2 = 2(3) – 7\)
\(1 = 6−7\)
\(1 =−1(False)\)
Try y = 1 :
\((2 – 1)^2 = 2(1) – 7\)
\(1 = 2−7\)
\(1 = -5 False\)
3. For x = 3:
\((3 – y)^2 = 3y -7\)
Try y = 3 :
\((3 – 3)^2 = 3(3) – 7\)
\(0=9−7\)
\(0 = 2 False\)
Try y = 1 :
\((3 – 1)^2 = 3(1) – 7\)
\(2^2 = 3 – 7\)
\(4 = -4( False )\)
4. For x = -1 and y = -1 :
\([ (-1 – (-1))^2 = -1(-1) – 7\)
\(0 = 1−7\)
\(0 = -6 \quad (\text{False})\)
5. For x = -1 and y = 1 :
\([ (-1 – 1)^2 = -1(1) – 7\)
\((-2)^2 = -1 – 7\)
\(4 = -8 \quad (\text{False})\)
6. For x = 2 and y = 4 :
\([ (2 – 4)^2 = 2(4) – 7\)
\((-2)^2 = 8 – 7\)
\(4 = 1 \quad (\text{False})\)
7.For x = 4 and y = 2:
\([ (4 – 2)^2 = 4(2) – 7\)
\(2^2 = 8 – 7\)
\(4=1(False)\)
✅Theonlyvalidsolutionsare:
\([(x,y)=(2,4)and(4,2)]\)
✅ Conclusion: No Valid Integer Solutions Exist
After a thorough and rigorous analysis, it turns out that there are no integer pairs (x,y) that satisfy the given equation.
✅ The correct conclusion is
\(\mathbf{\text{No integer solutions exist for } x^2 + y^2 + 7 = 3xy.}\)
Also Read About : Find All Integers n for which n^4+4^n is Prime – USAMO 2019
Khairun
