Befunge 98 - 17 11 9 8 bytes

'-:*b-.@ 

Similar to the old version, but I remembered about '

'-:* pushes 45, duplicates it, then squares it, producing 2025 b- subtracts 11 from it, resulting in 2014 .@ prints the result, then ends the program 

Interestingly, \$45^2-11\$ is the only pairing of numbers a,b where $$(a,b)∈[32,126]\times[10,15]\land a^2-b=2014$$ The significance of those sets is that \$[32,126]\$ is the set of printable ascii characters and \$[10,15]\$ is the set of easily accessible Befunge numbers. I found that pair with this python program:

for a in range(32,127): for c in range(10,16): if (a**2-c)==2014: print("%s,%s"%(a,c)) 

Or, if your interpreter supports unicode, then this works:

Befunge 98 - 5 bytes (4 chars)

'ߞ.@ 

It at least works on http://www.quirkster.com/iano/js/befunge.html with the following code (Befunge 93 - 6 bytes / 5 chars):

"ߞ".@ 

Old version

cdd**e-.@ 

computes the number, then prints it:

cdd pushes numbers to the stack so that it is this: 12,13,13 ** multiplies top three values of stack, which is now: 2028 e pushes 14 - subtracts the top two values of the stack, resulting in: 2014 . prints the numerical value @ end of program 

Older version:

"*'&("#;:a`j@a+,; 

Pushes the ascii values for 2014, -10. Then prints each after adding 10 to it.

Befunge-98 (FBBI), 17 11 9 8 bytes

'-:*b-.@ 

Try it online!

Similar to the old version, but I remembered about '

'-:* pushes 45, duplicates it, then squares it, producing 2025 b- subtracts 11 from it, resulting in 2014 .@ prints the result, then ends the program 

Interestingly, \$45^2-11\$ is the only pairing of numbers a,b where $$(a,b)∈[32,126]\times[10,15]\land a^2-b=2014$$ The significance of those sets is that \$[32,126]\$ is the set of printable ascii characters and \$[10,15]\$ is the set of easily accessible Befunge numbers. I found that pair with this python program:

for a in range(32,127): for c in range(10,16): if (a**2-c)==2014: print("%s,%s"%(a,c)) 

Or, if your interpreter supports unicode, then this works:

Befunge 98 - 5 bytes (4 chars)

'ߞ.@ 

It at least works on http://www.quirkster.com/iano/js/befunge.html with the following code (Befunge 93 - 6 bytes / 5 chars):

"ߞ".@ 

Befunge-98 (FBBI), 9 bytes

Old version:

cdd**e-.@ 

Try it online!

computes the number, then prints it:

cdd pushes numbers to the stack so that it is this: 12,13,13 ** multiplies top three values of stack, which is now: 2028 e pushes 14 - subtracts the top two values of the stack, resulting in: 2014 . prints the numerical value @ end of program 

Befunge-98 (FBBI), 17 bytes

Older version:

"*'&("#;:a`j@a+,; 

Try it online!

Pushes the ascii values for 2014, -10. Then prints each after adding 10 to it.

#Befunge 98 - 17 11 9 8 bytes

'-:*b-.@ 

Similar to the old version, but I remembered about '

'-:* pushes 45, duplicates it, then squares it, producing 2025 b- subtracts 11 from it, resulting in 2014 .@ prints the result, then ends the program 

Interestingly, 45²-11 is the only pairing of numbers a,b where $$(a,b)∈[32,126]\times[10,15]\land a^2-b=2014$$ The significance of those sets is that [32,126] is the set of printable ascii characters and [10,15] is the set of easily accessible Befunge numbers. I found that pair with this python program:

for a in range(32,127): for c in range(10,16): if (a**2-c)==2014: print("%s,%s"%(a,c)) 

Or, if your interpreter supports unicode, then this works:

#Befunge 98 - 5 bytes (4 chars)

'ߞ.@ 

It at least works on http://www.quirkster.com/iano/js/befunge.html with the following code (Befunge 93 - 6 bytes / 5 chars):

"ߞ".@ 

Old version

cdd**e-.@ 

computes the number, then prints it:

cdd pushes numbers to the stack so that it is this: 12,13,13 ** multiplies top three values of stack, which is now: 2028 e pushes 14 - subtracts the top two values of the stack, resulting in: 2014 . prints the numerical value @ end of program 

Older version:

"*'&("#;:a`j@a+,; 

Pushes the ascii values for 2014, -10. Then prints each after adding 10 to it.

Befunge 98 - 17 11 9 8 bytes

'-:*b-.@ 

Similar to the old version, but I remembered about '

'-:* pushes 45, duplicates it, then squares it, producing 2025 b- subtracts 11 from it, resulting in 2014 .@ prints the result, then ends the program 

Interestingly, \$45^2-11\$ is the only pairing of numbers a,b where $$(a,b)∈[32,126]\times[10,15]\land a^2-b=2014$$ The significance of those sets is that \$[32,126]\$ is the set of printable ascii characters and \$[10,15]\$ is the set of easily accessible Befunge numbers. I found that pair with this python program:

for a in range(32,127): for c in range(10,16): if (a**2-c)==2014: print("%s,%s"%(a,c)) 

Or, if your interpreter supports unicode, then this works:

Befunge 98 - 5 bytes (4 chars)

'ߞ.@ 

It at least works on http://www.quirkster.com/iano/js/befunge.html with the following code (Befunge 93 - 6 bytes / 5 chars):

"ߞ".@ 

Old version

cdd**e-.@ 

computes the number, then prints it:

cdd pushes numbers to the stack so that it is this: 12,13,13 ** multiplies top three values of stack, which is now: 2028 e pushes 14 - subtracts the top two values of the stack, resulting in: 2014 . prints the numerical value @ end of program 

Older version:

"*'&("#;:a`j@a+,; 

Pushes the ascii values for 2014, -10. Then prints each after adding 10 to it.

#Befunge 98 - 17 11 9 8 bytes

'-:*b-.@ 

Similar to the old version, but I remembered about '

'-:* pushes 45, duplicates it, then squares it, producing 2025 b- subtracts 11 from it, resulting in 2014 .@ prints the result, then ends the program 

Interestingly, 45²-11 is the only pairing of numbers a,b where

(a,b)∈[32,126]X[10,15]^a2-b=2014

The significance of those sets is that [32,126] is the set of printable ascii characters and [10,15] is the set of easily accessible Befunge numbers. I found that pair with this python program:

for a in range(32,127): for c in range(10,16): if (a**2-c)==2014: print("%s,%s"%(a,c)) 

Or, if your interpreter supports unicode, then this works:

#Befunge 98 - 5 bytes (4 chars)

'ߞ.@ 

It at least works on http://www.quirkster.com/iano/js/befunge.html with the following code (Befunge 93 - 6 bytes / 5 chars):

"ߞ".@ 

Old version

cdd**e-.@ 

computes the number, then prints it:

cdd pushes numbers to the stack so that it is this: 12,13,13 ** multiplies top three values of stack, which is now: 2028 e pushes 14 - subtracts the top two values of the stack, resulting in: 2014 . prints the numerical value @ end of program 

Older version:

"*'&("#;:a`j@a+,; 

Pushes the ascii values for 2014, -10. Then prints each after adding 10 to it.

#Befunge 98 - 17 11 9 8 bytes

'-:*b-.@ 

Similar to the old version, but I remembered about '

'-:* pushes 45, duplicates it, then squares it, producing 2025 b- subtracts 11 from it, resulting in 2014 .@ prints the result, then ends the program 

Interestingly, 45²-11 is the only pairing of numbers a,b where $$(a,b)∈[32,126]\times[10,15]\land a^2-b=2014$$ The significance of those sets is that [32,126] is the set of printable ascii characters and [10,15] is the set of easily accessible Befunge numbers. I found that pair with this python program:

for a in range(32,127): for c in range(10,16): if (a**2-c)==2014: print("%s,%s"%(a,c)) 

Or, if your interpreter supports unicode, then this works:

#Befunge 98 - 5 bytes (4 chars)

'ߞ.@ 

It at least works on http://www.quirkster.com/iano/js/befunge.html with the following code (Befunge 93 - 6 bytes / 5 chars):

"ߞ".@ 

Old version

cdd**e-.@ 

computes the number, then prints it:

cdd pushes numbers to the stack so that it is this: 12,13,13 ** multiplies top three values of stack, which is now: 2028 e pushes 14 - subtracts the top two values of the stack, resulting in: 2014 . prints the numerical value @ end of program 

Older version:

"*'&("#;:a`j@a+,; 

Pushes the ascii values for 2014, -10. Then prints each after adding 10 to it.