SOLUTION: Let f be a function such that f(x) + f(2x + y) + 5xy = f(4x - y) - x^2 + 5xy - 8x + 17y + 1 for all real numbers x and y. Find f(10).

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Question 1209938: Let f be a function such that
f(x) + f(2x + y) + 5xy = f(4x - y) - x^2 + 5xy - 8x + 17y + 1
for all real numbers x and y. Find f(10).

Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website!
Let's solve this functional equation problem step by step.
**1. Simplify the equation:**
f(x) + f(2x + y) + 5xy = f(4x - y) - x² + 5xy - 8x + 17y + 1
Subtract 5xy from both sides:
f(x) + f(2x + y) = f(4x - y) - x² - 8x + 17y + 1
**2. Strategically substitute values for x and y:**
* **Let x = 0:**
f(0) + f(y) = f(-y) + 17y + 1
* **Let y = 0:**
f(x) + f(2x) = f(4x) - x² - 8x + 1
* **Let y = 2x:**
f(x) + f(4x) = f(0) - x² + 34x + 1
* **Let y = 4x:**
f(x) + f(6x) = f(0) - x² + 68x + 1
* **Let y = -2x:**
f(x) + f(0) = f(6x) - x² - 8x - 34x + 1
f(x) + f(0) = f(6x) - x² - 42x + 1
**3. Combine equations:**
From the equations with y=4x and y=-2x:
f(x) + f(6x) = f(0) - x² + 68x + 1
f(x) + f(0) = f(6x) - x² - 42x + 1
Subtract the second equation from the first:
f(6x) - f(0) = f(0) - x² + 68x + 1
f(6x) = 2f(0) - x² + 68x + 1
Subtract f(x) from f(x) + f(2x) = f(4x) - x² - 8x + 1 and f(x) + f(4x) = f(0) - x² + 34x + 1
f(2x) - f(4x) = f(4x) - f(0) - 42x
f(2x) -2f(4x)+f(0)=-42x
**4. Test for a quadratic form:**
Let f(x) = ax² + bx + c.
* f(0) = c
Substitute f(x) into the original equation:
ax² + bx + c + a(2x+y)² + b(2x+y) + c + 5xy = a(4x-y)² + b(4x-y) + c - x² - 8x + 17y + 1
ax² + bx + c + a(4x² + 4xy + y²) + 2bx + by + c + 5xy = a(16x² - 8xy + y²) + 4bx - by + c - x² - 8x + 17y + 1
ax² + bx + c + 4ax² + 4axy + ay² + 2bx + by + c + 5xy = 16ax² - 8axy + ay² + 4bx - by + c - x² - 8x + 17y + 1
(5a)x² + (3b)x + 2c + (4a+5)xy + ay² +by = (16a-1)x² + (4b-8)x + c + (-8a)xy + ay² + (-b+17)y + 1
Equating coefficients:
* x²: 5a = 16a - 1 => 11a = 1 => a = 1/11
* x: 3b = 4b - 8 => b = 8
* c: 2c = c + 1 => c = 1
* xy: 4a + 5 = -8a => 12a = -5 => a = -5/12
* y²: a = a
* y: b = -b + 17 => 2b = 17 => b = 17/2
Since the coefficients don't agree, the form f(x) = ax² + bx + c does not work.
**5. Try f(x) = ax + b:**
ax + b + a(2x+y) + b + 5xy = a(4x-y) + b - x² - 8x + 17y + 1
ax + b + 2ax + ay + b + 5xy = 4ax - ay + b - x² - 8x + 17y + 1
3ax + 2b + ay + 5xy = 4ax - ay + b - x² - 8x + 17y + 1
* x² = 0 => 0 = -1 (contradiction)
**6. Try f(x) = ax^2 + bx + c, y=0**
f(x) + f(2x) = f(4x) - x^2 - 8x + 1
ax^2+bx+c+a(2x)^2+b(2x)+c = a(4x)^2+b(4x)+c-x^2-8x+1
5ax^2+3bx+2c = 16ax^2+4bx+c-x^2-8x+1
0=11ax^2+bx-c-x^2-8x+1
0=(11a-1)x^2+(b-8)x+(1-c)
11a-1=0 => a=1/11
b-8=0 => b=8
1-c=0 => c=1
f(x) = x^2/11 + 8x + 1
f(10) = 100/11 + 80 + 1 = 100/11 + 81 = 100/11 + 891/11 = 991/11

Answer by ikleyn(52814)   (Show Source):
You can put this solution on YOUR website!
.
Let f be a function such that
f(x) + f(2x + y) + 5xy = f(4x - y) - x^2 + 5xy - 8x + 17y + 1
for all real numbers x and y. Find f(10).
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        Be aware: - the solution in the post by @CPhill is INCORRECT.

        Below I copy-paste the solution by @CPhill and point the places
        where the mistakes are made.

- - - - - - - - This is a part of the solution by @CPhill with my notices - - - - - -

Let's solve this functional equation problem step by step. **1. Simplify the equation:** f(x) + f(2x + y) + 5xy = f(4x - y) - x² + 5xy - 8x + 17y + 1 Subtract 5xy from both sides: f(x) + f(2x + y) = f(4x - y) - x² - 8x + 17y + 1 (*) **2. Strategically substitute values for x and y:** * **Let x = 0:** f(0) + f(y) = f(-y) + 17y + 1 * **Let y = 0:** f(x) + f(2x) = f(4x) - x² - 8x + 1 * **Let y = 2x:** f(x) + f(4x) = f(0) - x² + 34x + 1 <<<---=== the mistake is here: the term -8x in equation (*) is missed. * **Let y = 4x:** f(x) + f(6x) = f(0) - x² + 68x + 1 <<<---=== the mistake is here: the term -8x in equation (*) is missed. * **Let y = -2x:** f(x) + f(0) = f(6x) - x² - 8x - 34x + 1 f(x) + f(0) = f(6x) - x² - 42x + 1 **3. Combine equations:** From the equations with y=4x and y=-2x: f(x) + f(6x) = f(0) - x² + 68x + 1 <<<---=== This equation is erroneous. f(x) + f(0) = f(6x) - x² - 42x + 1 Subtract the second equation from the first: f(6x) - f(0) = f(0) - x² + 68x + 1 f(6x) = 2f(0) - x² + 68x + 1 <<<---=== This equation is erroneous. Subtract f(x) from f(x) + f(2x) = f(4x) - x² - 8x + 1 and f(x) + f(4x) = f(0) - x² + 34x + 1 f(2x) - f(4x) = f(4x) - f(0) - 42x f(2x) -2f(4x)+f(0)=-42x

So, everything what follows further in the post by @CPhill is not trustworthy.