Prove that the number 40..09 (with at least one zero) is not a perfect square.

**!!** SelfReferential said:

(i didn't copy the first couple of lines fast enough, but this is the gist of it)

x^2 = 4*10^k + 9 = 40..09

x^2 - 3^2 = 4*10^k

See, remember: a-squared minus b-squared equals (a-b)*(a+b)

Let us postulate the number is squared and call it x-squared, where x is some integer. Then, as I said, we have x-squared minus 3-squared, which equals 4*(10^k), where k is a positive integer greater than 1

which means we've got (x-3)*(x+3) = 4*(10^k)

Since x is an integer and 3 is, too, that means (x+3) and (x-3) are both integers.

now see:

GCD ((x-3);(x+3) = GCD ((x+3);6) by Euclides' algorithm

since (x+3)-(x-3)=6

That means that (x-3) and (x+3) can't both divide on 5 as 6 does not divide on it

but they product equals 4*(10^k), then it must divide on 5^k

Then since they can't share the division powers, one of the two multipliers has to divide on 5^k all by themselves

So, one of the multipliers divides by 5^k, that means it equals (5^k)*t, where t is an integer

the difference between the two multipliers is 6

that means the other multipliers is either greater or less by 6

Thus, as one multiplier is (5^k)*t, so the other one is ((5^k)*t)±6

Their product, ((5^k)*t) x (((5^k)*t)±6), equals 4*(10^k), where k is an integer greater than 1

Dividing by 5^k, we get: t*(((5^k)*t)±6) = 4*(2^k) = 2^(k+2)

't' can be postulated positive:

(since *5^k)*t is one of the two multipliers' value, and the multipliers are x-3 and x+3, and as the original number is x-squared, you can postulate x = positive very well (it's square, so plus/minus makes no difference)

(and that means x>0, x+3>3>-3 and x-3>-3; both multipliers are greater than -3; one of them equals (5^k)*t, so (5^k)*t is greater than -3; since k is greater than 1, (5^k) is greater than 5^1, i.e. greater than 5.

that means that since (5^k)*t is greater than -3 and (5^k) is greater than 5, then t is greater than -3/5 = -0.6, which is greater than -1

so 't' is an integer greater than -1, that means it's either 0 or positive

If you put 0 in the equation (written at 15:27:28), you get 0 = 2^(k+2), which is impossible, so t is positive.

1) 't' is a positive integer, so greater-than-or-equal-to 1

k is a positive integer greater than 1, then it is greater-than-or-equal-to 2

Then 5^k is greater-than-or-equal-to 5^2 = 25

(5^k)*t is greater-than-or-equal-to 25t

Correction correction: [hold the (5^k) value] (5^k)*t is greater-than-or-equal-to (5^k)

Then, ((5^k)*t)±6) is greater-than-or-equal-to (5^k)-6

Since t is greater-than-or-equal-to 1, and ((5^k)*t)±6) is greater-than-or-equal-to (5^k)-6, which is greater-than-or-equal-to (5^2)-6=25-6=19>0, so

so t*(((5^k)*t)±6) is greater-than-or-equal-to 1*((5^k)-6)

Considering that t*(((5^k)*t)±6) = 4*(2^k) = 2^(k+2)

We get that 2^(k+2) is greater-than-or-equal-to (5^k)-6

Which is, as easily shown, not true for 'k' - integer greater than 2.

If you want detail for this as well: 2^(k+2) = 4*(2^k), postulate: 4*(2^k) is greater-than-or-equal-to (5^k)-6; divide by (2^k)>0 (no reverse), get: 4 is greater-than-or-equal-to (5^k)/(2^k) - 6/(2^k), which equals (5/2)^k - 6/(2^k)

(5/2)^k - 6/(2^k) grows when k grows (easy to show), then check" for k=2 (5/2)^2 - 6/(2^2) = 2,5 x 2,5 - 6/4 = 6,25 - 1,5 = 4,75 - greater than 4, then for all other k greater than 2 it's also greater than 4, so no solutions here! Yay!

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