An interesting task: Number is written w/ only 0, 3, 7; show there's no perfect squar

[happybeans]

problem

 

[Deleted member 4993]

Natural number is written with only 0, 3, 7. Show that a perfect square of this number does not exist.

I do not understand the problem as stated.

Every real number, except 0, can be multiplied by itself - hence for every real number (other than 0) there exists a perfect square for that number.

 

[Dr.Peterson]

Natural number is written with only 0, 3, 7. Show that a perfect square of this number does not exist.

I'd like to see the exact wording of the actual problem; but I wonder if it is this:

If a natural number is written (in base ten) using only the digits 0, 3, and 7, show that it is not a perfect square (that is, the square of a natural number).
​

If so, you might start by showing that no perfect square can have 3 or 7 in the units digit.

 

[Steven G]

Natural number is written with only 0, 3, 7. Show that a perfect square of this number does not exist.

If you square a number that ends in 1, what will the square of that number end in?

If you square a number that ends in 2, what will the square of that number end in?

If you square a number that ends in 3, what will the square of that number end in?

...

If you square a number that ends in 9, what will the square of that number end in?

What about a number that ends in 0?

How does this help?

To get further help, you need to answer these questions.

Thanks!

 

[JeffM]

So, I took a look at those number.
If I square a number that ends in 1, I will get numbers that end in 1.
If I square a number that ends in 2, I will get numbers that end in 4 (2^2 =4, 22^2=484, 16252^2=264127504)
I I square a number that ends in 3, I will get numbers that end in 9 (3^2=9, 373^2=139129)
..
I I square a number that ends in 0, I will get numbers that end in 0

Those numbers never end in 3 or 7.

To prove your point you needed to test all 10 digits.

The 10 digits can be expressed as follows.

\(\displaystyle \text {CASE I: } d = 5 \implies d^2 = 25.\)

Requires a 5.

\(\displaystyle \text {CASE II: } d = 5 \pm 1 \implies d^2 = 25 \pm (2 * 5 * 1) + 1 = 26 \pm 10 = 36 \text { or } 16.\)

Requires a 6.

\(\displaystyle \text {CASE III: } d = 5 \pm 2 \implies d^2 = 25 \pm (2 * 5 * 2) + 4 = 29 \pm 20 = 49 \text { or } 9.\)

Requires a 9.

\(\displaystyle \text {CASE IV: } d = 5 \pm 3 \implies d^2 = 25 \pm (2 * 5 * 3) + 9 = 34 \pm 30 = 64 \text { or } 4.\)

Requires a 4.

\(\displaystyle \text {CASE V: }d = 5 \pm 4 \implies d^2 = 25 \pm (2 * 5 * 4) + 16 = 41 \pm 40 = 81 \text { or } 1.\)

Requires a 1.

\(\displaystyle \text {CASE VI: }d = 5 - 5 \implies d^2 = 0^2 = 0.\)

Requires a zero.

Now you can do a proof by weak induction.

Prove that for any positive integer n, its square must contain a digit that is neither 0, 3, nor 7.

Note that in your base case, the possibility of zero has been eliminated by the requirement that n be a positive integer.

 

[HallsofIvy]

It is useful to note that (10a+ b)^2= 100a^2+ 20ab+ b^2= 10(10a^2+ 2ab)+ b^2 so the last digit of (10a+ b)^2 is the last digit of b^2.

 

[Dr.Peterson]

Since the problem as you stated it used the word "show", not "prove", it may not be necessary to do all the hard work of induction or whatever. There are ways to convince someone that it is true, without a formal proof. And these ways can often be turned into a formal proof if necessary, after you first get the main ideas clear.

I'll ask again, please copy the problem exactly as given to you, so we can be more sure what you need to do. It may also help if you tell us the context: is this for a course that focuses on proofs, or something else?

And then, please show what you've done so far.

 

[lookagain]

I do not understand the problem as stated.

Every real number, except 0, can be multiplied by itself - hence for every real number (other than 0) there exists a perfect square for that number.

Incorrect. Every real number can be multiplied by itself, including zero.

\(\displaystyle 0*0 \ = \ 0\)

I don't know what you were getting at.