Given a nonnegative integer \$n\$, determine whether \$n\$ can be expressed as the sum of two square numbers, that is \$\exists a,b\in\mathbb Z\$ such that \$n=a^2+b^2\$.
0 -> truthy 1 -> truthy 2 -> truthy 3 -> falsy 4 -> truthy 5 -> truthy 6 -> falsy 7 -> falsy 11 -> falsy 9997 -> truthy 9999 -> falsy Relevant OEIS sequences:
This is code-golf, so shortest answer as measured in bytes wins.
o=@(x)any(~mod(sqrt(x-(0:(x/2)^.5).^2),1)); taking advantage of vectorized code to search from 0 to sqrt(x/2), checking sqrt(x - i^2) is an integer with mod([],1)
o (so leaving it as an anonymous function definition), swapping (x/2) for just x, and outputting the other way around (so 1 for not sum of 2 squares) by removing ~ and changing to all... \$\endgroup\$ let a t=Seq.allPairs[0..t][0..t]|>Seq.tryFind(fun(f,s)->f*f+s*s=t) Straight-forward: create two lists from 0 to total, Seq.allPairs gets the Cartesian product between the two, and Seq.tryFind tries to find the pair that, when squared and added together, equals the total t.
f(n)=\prod_{a=0}^n\prod_{b=0}^n\{aa+bb=n:0,1\} The leading newline is necessary for the piecewise to paste properly.
Outputs 0 for truthy and 1 for falsey.
The ∏ trick didn't work, maybe due to the \{aa+bb=n:0,1\} piecewise. Avoiding it with sign(aa+bb-n)^2 or 0^{(aa+bb-n)^2} ended up longer.
int n->int=>{int a=0while a<=n{int b=0while b<=a if(n==a*a+b*b++)n=0a++}!n;} int n->int=>{ // Define a function taking 1 int and returning an int. Explicitely stating return type ends up faster than using `return` later. int a=0 while a<=n{ // Standard for(a=0; a<=n; a++) int b=0 while b<=a // for(b=0; b<=a if(n==a*a+b*b++) // We increment b as we check if a^2+b^2 to avoid needing blocks n=0 // Instead of returning, we just set n to 0, as n=0 is a known truthy state. a++ // And if n is zero, everything conveniently exists early. } !n; // if by the end, n is non-zero, this returns 0, or 1 otherwise. } 
{⍵=0:1⋄∨/0=1∣√⍵-×⍨⍳⌊√⍵} if input is n>=0, return 1 if n=0 or if n!=0, for some a in 1..√n, sqrt(n-a^2) is one integer. Could be this 15 if one can apply "0 is true", "!=0 is false" and input>0
{×/1∣√⍵-×⍨⍳⌊√⍵} Test:
(0,⍳100)/⍨ {⍵=0:1⋄∨/0=1∣√⍵-×⍨⍳⌊√⍵}¨0,⍳100 0 1 2 4 5 8 9 10 13 16 17 18 20 25 26 29 32 34 36 37 40 41 45 49 50 52 53 58 61 64 65 68 72 73 74 80 81 82 85 89 90 97 98 100 
PODPROGRAM:F(N) **)LINIIENT(A*2+B*2-N) POWTORZ:A=0(1)N POWTORZ:B=0(1)N WROC Outputs 0 carriage returns for falsy and a positive number of carriage returns for truthy. An easy way to check how many were printed is by using this command: ./filename | grep $'\r' -c.
CZYTAJ:N **1)LINIIENT(A*2+B*2-N) POWTORZ:A=0(1)N POWTORZ:B=0(1)N STOP1 KONIEC 
f=n=>(m=n)?[...Array(++m*m).keys()].some(i=>(i%n)**2+(~~(i/n))**2==n):0 Not great but might as well post it since I spent an hour trying.
{for(;i<=$1;i++)for(j=-1;j++<$1;)j^2+i^2-$1||g=1}$0=g It's not fast when you get to bigger numbers as it's exhaustive and doesn't short when it finds one.
{for(;i<=$1;i++) # all numbers up to input for(j=-1;j++<$1;) # all numbers up to input j^2+i^2-$1||g=1} # it's not a good sum or set output to 1 $0=g # prints a one if a good sum is found 