help with a proof

[becko]

I cannot find a way to prove this:

 

A function r(n), where n is a positive integer, gives the number n with its digits inverted. Prove that for positives integers a and b, the two numbers

4(a^2) + r(b) and 4(b^2) + r(a) cannot be simultaneously perfect squares.

 

Can anyone help me?

[D H]

Show your work so far. People will be much more willing to help you after you have shown your attempts at solving the problem.

[becko]

I haven't had any advance. I'd appreciate any ideas. Maybe if we assume that one of the numbers is a perfect square, so that

4(a^2) + r(b) = k^2, where k is a positive integer, then as a consequence,

4(b^2) + r(a) shouldn't be a perfect square. Or if we assume that both numbers are perfect squares, we would surely arrive at some contradiction. I have tried to do both ways, but I have got nowhere. All I have are some crumpled papers with nothing valuable in them. Maybe someone can show me how to prove this?

[psi20]

I haven't got much from it either. Suppose that (2a)^2 + r(b) = k^2 and k is a positive integer. Then

 

k^2 - r(b) must be a multiple of 4.

 

k and k^2 are odd if and only if r(b) is odd.

 

k and k^2 are even if and only if r(b) is even. If k is even, r(b) must then be a multiple of 4 since k^2 is a multiple of 4 (k is even, k^2 must be even evener) and k^2 - r(b) is a multiple of 4.

 

If r(b) is odd, it has to be a certain kind of odd number. I haven't looked at this in depth. (2a)^2 is an even number. Again, if r(b) is odd, then k^2 is odd. Write out the first 10 squares. Subtract an even square (2a^2) from an odd square (k^2). There's a pattern to what kind of odd number r(b) can be.

 

0 1 4 9 16 25 36 49 64 81 100

 

k^2 - (2a)^2 = r(b)

Here's a list of odd squares minus even squares.

1, 5, 9, 13, 17... Something like 4x+1 or 4*1x+1^2

9, 21, 33, 45... 12x+9 or 4*3x+3^2

25, 45, 65...20x+25 or 4*5x+5^2

49, 77...28x+49 or 4*7x+7^2

 

I'm thinking that r(b) must fit this pattern.

 

Departing from this train of thought, (k+2a)(k-2a) = r(b).

 

Similarly, if (2b)^2 + r(a) = m^2, then the stuff from above similarly applies.

 

I've tried to find a contradiction by looking at the sum, difference, and product of k^2 and m^2. I've looked at just the evenness and oddness of the sum, difference, and product. I don't think it works, though, because r(b) and r(a) is even or odd like k and m.

 

I don't know if this will help. Perhaps it will, perhaps it won't.

[psi20]

If r(b) is even, it too has to be a specific type of number.

 

r(b) must be of the form y(4x+y) where x and y are positive integers.

r(a) must be, too. Again, I don't know if this will help.

 

*On the list of integers above, I shouldn't have listed 0 since we're talking about positive integers.

[psi20]

b = r(r(b)) and a = r(r(a)) so they both must be of the form y(4x+y).

 

k^2 - m^2 must be of the form y(2x+y). These y's and x's are different than the ones above. Might want to use different letters.

 

I don't know, but try looking into the method of infinite descent.