$\pmod 8$ qed

wording has been corrected thanx to Quidditch, it asks for non modulo solution

without modulo $a^2+b^2=6+8c$ implicating (a,b) both odd or both even if both even $4(a')^2+4(b')^2 = 8c+6$ $4c+3 =2(a)^2+2(b)^2$ contradiction since 3 is odd if a ,b odd $(2a'+1)^2+(2b'+1)^2=6+8c$ $4(a')^2+4(a')+1+(4b')^2+(4b')+1 = 6+8c$ $4+8c = 4a^2+4(a')+4(b')^2+(4b')$ $1+2c = a^2+a+b^2+b$ contradiction since RHS is due to a(a+1) is two consecutive number meaning one of them should be even same as b^2+b