What's so special about 1987?

Number Theory Level 4

$(m^{1987}-m+1987)^{m^{1987}-m+1987} = (n^{7891}-n+7891)^{n^{7891}-n+7891}$

Find the number of solutions to the Diophantine equation above?

Hint: Work in modular arithmetic using a small mod.

Details and Assumptions

If there are infinitely many solutions, input $-1$ .

1 solution

Jake Lai
Mar 1, 2015

We work in mod 7 to show that there are no solutions.

We know $\varphi(7) = 6$ . Hence, taking mod 7 on both sides' bases,

$m^{1987}-m+1987 \equiv m^{c_{1}\varphi(7)+1}-m+7c_{2}-1 \equiv -1 \pmod{7}$

$n^{7891}-n+7891 \equiv n^{c_{3}\varphi(7)+1}-n+7c_{4}+2 \equiv 2 \pmod{7}$

The index of the LHS is always odd; as for the RHS, $2^{k} \equiv 1,2,4 \pmod{7}$ for all $k \in \mathbb{N}$ , so LHS will never be congruent to RHS.

Hence, there are $\boxed{0}$ solutions to the Diophantine equation.

Why did you specifically choose mod 7 ?

A Former Brilliant Member - 6 years, 3 months ago

Honestly, I have no idea.

Jake Lai - 6 years, 3 months ago

As pointed out by Calvin Lin, if LHS is $a^{a}$ and RHS $b^{b}$ (as in the Diophantine) then $a = b$ . Then, proceed with parity.

Jake Lai - 6 years, 3 months ago