A379925 Numbers k for which nonnegative integers x and y exist such that x^2 + y^2 = k and x + y is a square.

0, 1, 8, 10, 16, 41, 45, 53, 65, 81, 128, 130, 136, 146, 160, 178, 200, 226, 256, 313, 317, 325, 337, 353, 373, 397, 425, 457, 493, 533, 577, 625, 648, 650, 656, 666, 680, 698, 720, 746, 776, 810, 848, 890, 936, 986, 1040, 1098, 1160, 1201, 1205, 1213, 1225, 1226

Offset: 1

Author

Felix Huber, Jan 25 2025

Keywords

• easy
• nonn

Comments

Numbers k for which exists at least one solution to k = x^2 + (z^2 - x)^2 in integers x and z with x >= 0 and z >= sqrt(2*x).

Subsequence of A001481.

Examples

			10 is in the sequence because 10 = 1^2 + 3^2 and 1 + 3 = 2^2.
81 is in the sequence because 81 = 0^2 + 9^2 and 0 + 9 = 3^2.
		

Links

• Felix Huber, Table of n, a(n) for n = 1..10000
• Wikipedia, Sum of two squares theorem

Crossrefs

Cf. A000161, A000290, A001481, A004018, A025284-A025320, A091072, A125110, A118882, A125022, A380072, A380073, A380074.

Programs

Maple

# Calculates the first 10005 terms.
A379925:=proc(K)
    local i,j,L;
    L:={};
    for i from 0 to floor(sqrt((K+1)^2)/2) do
        for j from 0 to floor(sqrt((K+1)^2/2-i^2)) do
            if issqr(i+j) then
                L:=L union {i^2+j^2}
            fi
        od
    od;
    return op(L)
end proc;
A379925(1737);

PARI

isok(n)=my(x=0, r=0); while(x<=sqrt(n) && r==0, if(issquare(n-x^2) && issquare(x+sqrtint(n-x^2)), r=1); x++); r; \\ Michel Marcus, Feb 10 2025

Formula

k = m^(4*j) is in the sequence for nonnegative integers m and j (not both 0) because x = 0 and z = m^j is a solution to m^(4*j) = x^2 + (z^2 - x)^2.