Prove that if $a$ is a natural number, then there exist two unequal natural numbers $k$ and $l$ for which $a^k-a^l$ is divisible by 10.
THIS IS WHAT I DID TO ANALYZE THE PROBLEM:
(1) For the difference of two numbers $x$ and $y$ to be divisible by 10,their "units digits" must be the same.
(2) I generated a table for some sample values of $a$, $k$, and $l$. By way of conjecture, I noticed that $a^k-a^l$ is divisible by 10 when $l = k + 4$ for a number of combinations of those variables.
(3) Substituting $k+4$ for $l$, I factored the original statement into $a^k(1-a^4)$.
(4) Generating a table for (3), I noticed four cases:
(A) $a$ is a multiple 5. Then $1-a^4$ has a units digit of four and $a^k$ has a unit digits of five. Multiplying the two always results in a number with a units digit of zero because four is even.
B) $a$ ia a multiple of 10. Then $1-a^4$ has a units digit of nine and $a^k$ has a units digit of zero. Multiplying the two always results in a number with a units digit of zero.
C) $a$ is an odd number not equal to five. Then $1-a^4$ has a unit digits of zero, so multiplied by any $a^k$ results in a number with a units digit of zero.
D) $a$ is an even number not equal to ten. Then $1-a^4$ has a units digits of five and $a^k$ is always even because an even number to a power is always even. Multiplying the two always results in a number with a units digit of zero
(5) In all cases the result has a units digit of zero, so they are all divisible by 10. These combinations of $k$ and $l$ map into a function $a^k-a^l$ divisible by ten.
HOWEVER, THE QUESTION IS THIS: Is all this analysis overkill? Does the analysis in anyway have a bearing on determine how many pigeons and how many pigeonholes there are? The aim of the problem is to use the pigeonhole principle to prove that a function is not injective so $k$ and $l$ can be two different natural numbers.