Squares! Sum of squares!

Number Theory Level 5

Let $a$ , $b$ , and $c$ be positive integers that satisfy the equation $a^{2}+98b^{2}+114c^{2}=1111111111$ (There are 10 1's) Find the sum of all possible values of $a$ , $b$ , and $c$ .

The answer is 0.

2 solutions

Sean Ty
Jun 20, 2014

Alright, the thing here isn't the number of $1$ 's, nor is it the coefficients of $b$ and/or $c$ .

We first rearrange the equation to $a^{2}+2b^{2}+96b^{2}+2c^{2}+112c^{2}$ = $1111111 \times 1000 +111$ .

We then apply modulo 8 on both sides. $a^{2}+2b^{2}+2c^{2}\equiv 7 \pmod{8}$ .

$a^{2}\equiv 0,1,4 \pmod{8}$

$2b^{2}\equiv 0,2 \pmod{8}$

$2c^{2}\equiv 0,2 \pmod{8}$

We can see that there is no way of achieving the equation of $a^{2}+2b^{2}+2c^{2}\equiv 7 \pmod{8}$

Thus, there are $\boxed{0}$ integer solutions.

You put that $a^2 \equiv 0, 1, 4, 9 \pmod{8}$ , but $9 \equiv 1 \pmod{8}$ , so it is a little repetitive.

I will now show my almost exactly same solution: Let us consider the parity. Since both $98b^2$ and $114c^2$ are even, $a$ must be odd. When dealing with squares of odd numbers, mod 4 and mod 8 are powerful, since for any odd number $n$ , $n^2 \equiv 1 \pmod {4,8}$ . We proceed with mod $8$ : $a^2+2b^2+2c^2\equiv 1+2b^2+2c^2 \equiv 7 \pmod{8} \implies 2b^2 + 2c^2 \equiv 6,$ which is impossible since for any integer $m$ , $2m^2 \equiv 0, 2 \pmod{8}$ .

Hahn Lheem - 6 years, 10 months ago

Oops, edited!

Sean Ty - 6 years, 10 months ago

Saranraj Gururaj
Aug 13, 2014

There is a simple logic, if a, b, c is a solution then -a,-b,-c is also a solution . So sum of all the solution cancel anyway(if exist).

Squares! Sum of squares!

The answer is 0.

2 solutions

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