MHB Find all Possible Values of f(3) from f(1)=10, f(5)=206

[Albert1]

Given a polynomial function with non-negative integer coefficients,
if f(1)=10,f(5)=206, find all the possible values of f(3) =?

 

[kaliprasad]

Given a polynomial function with non-negative integer coefficients,
if f(1)=10,f(5)=206, find all the possible values of f(3) =?

Spoiler

The polynomial can be maximum a cubic polynomial because $5^3 < 206$ and $5^4 = 625 > 206$
polynomial is not constant polynomial because $f(1)$ and $f(5)$ are not same
now let us consider linear quadatic and cubic polynomials
case 1
linear Let $f(x) = ax+b$
$f(1) = a + b = 6$ and $5a + b = 206$ gives $a = 40$ and $b= - 1$ invalid solution ( does not meet criteria)
case 2
now consider qudartic
$f(x) = ax^2+bx+c$
$f(1) = 10 => a + b+ c = 10\cdots(1)$
$f(5) = 206=> 25a + 5b+ c= 206\cdots(2)$
subtracting (1) from (2) we get $24a+4b= 196$ or $6a + b = 49$ giving solution $a = 8, b= 1$ in the range so we have from(1)
$c=1$

$f(x) = 8x^2+x + 1$ and f(3) = 76
this is one value
case 3)
now consider cubic
$f(x) = ax^3+ bx^2 + cx+d$ and a cannot be $>1$ as $f(5)$ becomes greater than 249
so $a=1$
we have
$f(x) = x^3 + bx^2+ cx + d$
$f(1) = 1 + b + c + d = 10$ or $b+c+d = 9\cdots(1)$
$f(5) = 125 + 25b + 5c + d = 206$ or $25b+5c +d = 81\cdots(2)$
subtract (1) from (2) to get
$24b+4c= 72$ or $6b+c = 18$ giving $b = 2,c = 6$ (which gives d = 1 from (1)) or $b=3,c = 0$ which gives $d=6$
giving 2 sets solutions
$f(x) = x^3 + 2x^2 + 6x + 1$ giving $f(3) = 64$
and
$f(x) = x^3 + 3x^2 + 1$ giving $f(3) = 60$

so we have 3 possible values for $f(3)$ 60, 64, 76