A Tricky Diophantine Equation

Number Theory Level 5

Find the last three digits of the sum of all positive integers $n \leq 2014$ for which there exist integers $a, b, c, d, e$ (not necessarily distinct) such that $a^2+b^2+c^2+d^2+e^2= \left( 3^n + 1 \right) ^2 \left( 2 \cdot 3^{2n} + 5 \right) .$

The answer is 105.

1 solution

Siddhartha Srivastava
Mar 14, 2014

Let $e = 0$ . We have $a^2 + b^2 + c^2 + d^2 = (3^n+1)^2(2 \times 3^{2n} + 5)$

The RHS is a natural number. By Lagrange's four square theorem, any natural number can be written as the sum of four squares. Therefore,there exist integers $a,b,c,d,e$ which satisfy the equation for all n.

Thus the sum = $\frac{2014\times2015}{2} = 2029105$

Answer = $\boxed{105}$

I thought it might be fun to write out a general solution.

$a=(3^n+1) \times 3^n$

$b=(3^n+1) \times 3^n$

$c=(3^n+1) \times 2$

$d=(3^n+1)$

$e=0$

Stephen Tosh - 7 years, 2 months ago

That explains the 5 there. And I'm pretty sure that was the solution he was looking for.

Nice Solution.

Siddhartha Srivastava - 7 years, 2 months ago

Nicely done!

Grace Doughty - 7 years, 2 months ago

It works even if we add the restriction that none of $a,b,c,d,e$ can be zero:

$a = (3^n+1)^2$

$b = (3^n+1)(3^n-1)$

$c = d = e = 3^n + 1$

Ariel Gershon - 6 years, 8 months ago

A Tricky Diophantine Equation

The answer is 105.

1 solution

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