Bhaskara Set Problem-1 (2004)

The equation can be rewritten as $a^2+b^2=8c+6$ . Taking $\pmod{8}$ on both sides we see that $a^2+b^2\equiv 6\pmod{8}$ .

However, quadratic residues $\pmod{8}$ are $0,1,4$ so the sum of two squares can never be $6\pmod{8}$ . Thus our answer is $\boxed{0}$ .

I have two solutions

$1)$ We can write the equation ${ a }^{ 2 }+{ b }^{ 2 }-8c=6$ to ${ a }^{ 2 }+{ b }^{ 2 }=8c+6$ , and this equation is equal ${ a }^{ 2 }+{ b }^{ 2 }\equiv 6(mod\quad 8)$ . We know that all the numbers $k^{ 2 }\equiv 0,1,4(mod\quad 8)$ so based on this, the $6\ (mod\quad 8)$ can´t be reached.

$2)$ Now. see that ${ a }^{ 2 }+{ b }^{ 2 }\equiv 2(mod\quad 4)$ so $a,b$ must be odd. Then, we can write $a,b$ as $a=2{ a }_{ 1 }+1,b={ 2b }_{ 1 }+1$ . So, the equation can be written as $(2{ a }_{ 1 }+1)^{ 2 }+({ 2b }_{ 1 }+1{ ) }^{ 2 }=8c+6$ , and this equation as $4{ a }_{ 1 }^{ 2 }+4{ a }_{ 1 }+4{ b }_{ 1 }^{ 2 }+{ 4b }_{ 1 }+2=8c+6\\ =4({ a }_{ 1 }^{ 2 }+{ a }_{ 1 }+{ b }_{ 1 }^{ 2 }+{ b }_{ 1 })=8c+4\\ ={ a }_{ 1 }({ a }_{ 1 }+1)+{ b }_{ 1 }({ b }_{ 1 }+1)=2c+1$

, but this is a contradiction because ${ a }_{ 1 }({ a }_{ 1 }+1)$ and ${ b }_{ 1 }({ b }_{ 1 }+1)$ are even. So, the answer is $\boxed { 0 }$ .

Note that $6+8c$ is $2$ mod $4$

The sum of squares is $2$ mod $4 \Rightarrow$ Both numbers are odd.

Therefore, $a,b$ are odd. So, $a^2=8x+1$ , $b^2=8y+1$

Therefore, dividing L.H.S. and R.H.S. by 4, we get $2x^2+2y^2+2c=1$

L.H.S. is even, R.H.S. is odd, thus there is no integral solution

P.S. Are you writing the Bhaskara contest in AMTI NMTC this year?