Three Squares Make another Square?

$\huge 1+4! -9 = 16$

Here, $!$ refers to factorial .

$1 \_ 4 \_ 9 = 16$ $(1+4)^{2}-9=16$ $(5)^{2}-9 = 16$ $25-9=16$ $16=16$

nice one: $\huge{1+\Gamma(4)+9=16}$ the gamma function is being used here.

How about this: 1<<(4%9) where << is the left shift operator.

1 + !4 - 9 = 16 is perfect, but if I'm also allowed to play with exponents, I would just do:

1 * 4^2 * 9^0 = 16

(-1-4+9)^2

-1 -4 +9 = sqrt(16)

I am curious what if 14/9=1.6 do you guys think it s allowed

The best solution is.
1+(4!-9)=16

4^(√9 - 1) = 4^2 = 16

1+(4^2)-(9^0)

(1*(4%9))^2 = 16,

where % = modulo

1+4!-9 = 16

-1-(4^3-9^2)=16

1^49 = 1^6

(1+4)^2 -9

1^2-(squar root of 4)×9

2-(2)×9

2-18=16

We're allowed to use exponents, right?

(9-1) x 4^1/2=16

Boy I do the easy route 1 + 4 + 9 ( does not equal sign ) 16

149>16 Evaluates to true. I win.

$1 + 4 - 9 = -\sqrt{16}$

P.S. How d

4{1+(9^0.5)}

4^([root 9] - 1)

$(\sqrt{9} + 1) \times 4 = 16$ , where $\sqrt{9}$ denotes the principal square root of 9

(\sqrt{9} +1)4

1(-4+9)=-1+6

((1+4) - 9)^2

$\huge -(1+ 4^3 - 9^2 )$

1+(4)!-9=16

-1 + cube(sqrt(4)) + 9

1,4, and 9 are all square numbers. So, it is possible to make a 16 because it also a perfect square.

Is 1x4squaredx9 to the 0 power allowed?

1 + 4^{2} - 9^{0}

f(x)=0 g(x)=16

f(1)+f(4)+g(9)=16

1^0+4^2-9^0

It says any function: we define (for example) f(x)=x+2, thus: f(1+4+9)=16.

$-(1+4)+9=\sqrt{16}$

Let $f(x) = 0$ and $g(x) = 16$

$f(1) + f(4) + g(9) = 16$

If any function can be used;

Define a function f(x) = 0

Then, f(1)+f(4)+f(9) = f(16)

Or

define g(x) = 14

1+4+9=g(16)

My solution was √(1 * 4 * 9)=1*6

(1+4^2) \ times 9^0

1x(4^2)+(9^0)

(1+4)^2 - 9 =16

4! - (9-1) = 16

(1) (4^1/2)

((1+4)^2)-9

1 + 4^2 - 9^0=16

1+16-1=16

16=16

1 + 4^{2} - 9^{0 }

1+4^2-9^0 = 1+16-1 = 16

(1+4)^2 - 9

Square of (4+1) - 9 =16

1+(4^2)-(9^0)=16

-1 -4^3 + 9^2 = 16

1+4sqr - 9 to the 0 power 1+ 16 -1 =16

1square 2square 3square=4square thats my logic

(1) (4^2) (9^0)=16

1 4 - 9 = -1 + 6

1 + 4^2 - 9^0

(1+4)^2 - 9 =9

4!-(9-1) >> 24-8 = 16

Square root of 9=3 3+1x4

Sol 1 4^(9^0.5 - 1) =4^(3-1) =4^2 =16

Sol 2 (1+4)^2 - 9 =5^2 - 9 =25 - 9 = 16

Sol 3 1+4! - 9 =1+24 - 9 = 25 - 9 =16

(1+4) squared - 9 = 16

1×(p[4]+9)=16 p[n] refers to nth prime.

9 squared - 4 cube - 1 = 16

(1^0)×(4^2)×(9^0)

4=2+2

((2 9)-2) 1=16

(18-2)*1=

16*1=16

(−1+(√4+√9))^(2)

1^0+4^2-9^0

(1)^2+(4)^2-(9)^0 = 16

-1-4^3+9^2

9^2-sqrt(4)=16

(-1-4+9)=_/16

(1+4)^2-9=16

Same as Anish. 4!=24, so 1+24-9=16.

1+4!-9=16 is the answer!!!

(1+√9)×4=16

(1+4)^2-9=16 ^2 is squared

(Sqr rt 9 +1)*4

1^0 X 4^4 X 9^0 1x16x1= 16

1 + 4! - 9

(1^2+4^2)-9^0=16

4! + 1 - 9 = 16 4! = (4)(3)(2)(1) = 24 + 1 = 25 - 9 = 16

(1+4)^2-9=25-9=16

(9-1)*sqrt{2}

1 + 4! - 9 = 16

I almost forgot about factorial numbers xD

Note: this is not a solution

$\huge{1+4+9\neq 16}$

(1+4!)-9=16 Please note that parentheses are used for better understanding of the problem.

(-1-4+9)^2 = 16

( Sqr.(9)+1)* 4= 16

$1 + 4^{2} - 9^{0}$

(1+4)^2-9 =5^2-9 =25-9=16

(1 + 4!) - 9 = 16 or (1 + 4)^2 - 9 = 16

1+(4!-9)=16