How to show that for each integer n≥0, 1+10^(231+462n) is divisible by the squares of at least two different primes?

It's divisible by 11^2.

Since 231+462n is odd, the expression is the sum of two odd powers: 1^(231+462n) + 10^(231+462n). So you can factor it as (1 + 10)(1 - 1(10) + 1(10)^2 + ... + 10^(230+462n))

Mod 11, every term of the 2nd factor is = 1, so the entire expression is equal to 1 X the number of terms, which is 231 + 462n = 11(21 + 42n), so it is divisible by 11. Therefore the number is divisible by 11^2.

The key here is that the exponent is divisible by 11, so that you have a multiple of 11 terms = 1 mod 11. The numbers are way to big to actually calculate. Here's a simpler example:

1000000000000000000000000000000001 = 1 + 10^33 =

8 264 462 809 917 355 371 900 826 446 281 times 121

It's also divisible by 7^2.

Divide the number by (1 + 10^3) = 1001 = 143x7

You get (1 - 10^3 + 10^6 - 10^9 .... + 10(231 + 462n - 3))

Each of those terms = 1 mod 7. There are (231 + 462n)/3 = 77 + 154n = 7(11+22n) terms. Therefore the sum of them = 0 mod 7, and is therefore divisible by 7. That gives us a second factor of 7, so the expression is divisible by 7^2.

The key here is that the 231+462n is divisible by both 3 and 7. Again it's too large to calculate, but a simpler example illustrates it:

1000000000000000000001 = 10^21 + 1 =

20408163265306122449 x 49

Fairly tough problem! It doesn't take any really advanced math to solve this, just algebra and modular arithmetic, but I found it a bit tricky to dig out the solution.