The two consecutive odd positive integers are 11 and 13.

Here’s a quick way to see why: Let the first odd positive integer be xxx, so the next consecutive odd integer is x+2x+2x+2.

Their squares are x2x^2x2 and (x+2)2(x+2)^2(x+2)2, and their sum is given as 290, so:

x2+(x+2)2=290x^2+(x+2)^2=290x2+(x+2)2=290

x2+x2+4x+4=290x^2+x^2+4x+4=290x2+x2+4x+4=290

2x2+4x+4=2902x^2+4x+4=2902x2+4x+4=290

2x2+4x−286=02x^2+4x-286=02x2+4x−286=0

Divide by 2:

x2+2x−143=0x^2+2x-143=0x2+2x−143=0

This factors as:

(x−11)(x+13)=0(x-11)(x+13)=0(x−11)(x+13)=0

So x=11x=11x=11 or x=−13x=-13x=−13, and only x=11x=11x=11 is a positive odd integer.

Thus the two consecutive odd positive integers are 11 and 13 , and:

112+132=121+169=29011^2+13^2=121+169=290112+132=121+169=290

which matches the given condition.