#1 Measure you Calibre

Number Theory Level 4

$\large 5 \times 3^m + 4 = n^2$

Consider all possible integers $n \ge 0$ such that the equation above holds true for some corresponding integer $m \ge 0$ . Find the sum of all such $n$ .

Check your Calibre

The answer is 10.

1 solution

Tapas Mazumdar
Mar 5, 2017

$(5 \times 3^m) + 4 = n^2 \\ \implies (5 \times 3^m) = (n+2)(n-2)$

Notice that LHS has two coprime factors. In the RHS $\text{gcd}(n+2,n-2) = \text{gcd}(4,n+2)$ (as $\text{gcd}(a+b,a-b) = \text{gcd}(a+b,2b)$ ), here, $n+2$ should be odd which means that gcd is 1, hence RHS must contain both factors coprime as well.

Hence

$n+2 = 5 \implies n=3$

$n-2 = 5 \implies n=7$

We can check that

$n=3 \to 5 \times 3^m = 5 \implies m = 0 \ (m : m \in \mathbb{Z}, m \ge 0)$

$n=7 \to 5 \times 3^m = 45 \implies m = 2 \ (m : m \in \mathbb{Z}, m \ge 0)$

Thus the sum is $3+7 = 10$ .

WOW VARY CUL

Sahil Silare - 4 years, 3 months ago

Thanks pal.

Tapas Mazumdar - 4 years, 3 months ago

To fill in the gap, notice that $gcd(n+2, n-2) = \gcd (4, n+2)$ is a factor of 4. Since $n+2$ is odd, hence the gcd = 1, thus they are coprime.

Note: You also have to check that the other factor is a power of 3.

Calvin Lin Staff - 4 years, 3 months ago

Thanks. I've made the rightful changes accordingly.

Tapas Mazumdar - 4 years, 3 months ago

Is there a proof to $gcd(a-b,a+b) = gcd(a+b,2b)$ ?

Vishal Yadav - 4 years, 2 months ago

Can you show that $\gcd (n,m) = \gcd (n, n+m) = \gcd ( n, m-n)$ ?

More generally, for any integer X, $\gcd (n,m) = \gcd (n, m+Xn)$ .

Check out the greatest common divisor wiki for more information :)

Calvin Lin Staff - 4 years, 2 months ago

#1 Measure you Calibre

The answer is 10.

1 solution

0 pending reports