Please someone disprove that such numbers exists for any n digits, n positive integer

[steel linden]

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[undone raven]

Because I think proving this fact is kinda too hard

and also I don't think that it's true in the first place

Please someone disprove that such numbers exists for any n digits, n positive integer

[real oasis]

Is there any such 2-digit number?

[undone raven]

Idk

[real oasis]

Let 10a + b be such a 2-digit number with 1 <= a <= 9 and 0 <= b <= 9, hence a^2 + b^2 = 10a + b.
Suppose b in {0, 1, 5, 6}. Then b - b^2 is a multiple of 10, and thus so is a^2 = b - b^2 + 10a. Contradiction.
Suppose b in {2, 4, 7, 9}. Then b - b^2 has a remainder of 8 when divided by 10, and thus so does a^2. Contradiction.
Suppose b in {3, 8}. Then b - b^2 has a remainder of 4 when divided by 10, and thus so does a^2.
2^2 + 3^2 = 13, not 23.
8^2 + 3^2 = 73, not 83.
2^2 + 8^2 = 68, not 28.
8^2 + 8^2 = 128, not 88.
Hence a contradiction arises. There is no such 2-digit number.

@undone raven

[undone raven]

@real oasis That's a nice proof. Although it's correct I think you should have also say that 13 is not 23 NOR 32 since the order of the sum 2^2 + 3^2 = 3^2 + 2^2. The same thing for
73 != 83 nor 38
68 != 28 nor 82
128 != 88 nor 88 (redundant here)

Other than this, thanks for your help

Although, do you think is there another way to prove in thcase of 3 digits other than your brute force way?

[real oasis]

I think I rejected unit digit 2 elsewhere, didn't I?

[undone raven]

no

its in the case {2,4,7,9}

[real oasis]

Suppose b in {2, 4, 7, 9}. Then b - b^2 has a remainder of 8 when divided by 10, and thus so does a^2. Contradiction.

Exactly

So I don't have to consider 32 anymore

[undone raven]

aaaaaaaaaaa

you're right...

But how about in the case of 3 digits?

Is there another way of proving that there isn't any without the brute force way

[real oasis]

I thought the job was done, that we only had to disprove that for any natural n, there is some such number

I am not really intrigued enough to continue brute-forcing for obvious reasons
Let me sleep on it (albeit not spend extra effort) and give you an update if any idea pops up

[undone raven]

ok