prove:x squared congruent to 45 modulo 100 has no integer solutions?

Answer 1

What you want to prove is that no perfect square, expressed in base 10, can end in 45.

To prove this:
- Ask yourself: If a square ends in 5, what is the last digit of its square root?
- Then ask: If a number ends in that digit, can its square end in 45?

Each of these questions is relatively easy to answer, and answering them will give you what you need to write the proof.

Answer 2

Basically, you're asking if, after squaring a number, the resultant perfect square can end in the digits 45. No, it can't.

First, the number has to end in 5, since these are the only numbers that, when squared, give a number ending in 5.

So, call your number (10a+5). Squaring this number gives 100a^2 + 100a + 25. The first two terms are divisible by 100, therefore the last two digits (i.e. the value if taken mod 100) must be 25.

Answer 3

If true, then x^2 = 100*i + 45 = 5*(20*i + 9) where i is an integer.

However, for this to be true then 20*i + 9 must be divisible by 5. That is impossible since 5 divides 20*i and 5 does not divide 9.

Answer 4

If x^2 ends in 5 then x has to end in 5, easy to check. So you only need to check the last two digits of the squares of 5, 15, 25,...,95 to be able to say there is no solution.