Prove that an integer with digits '1' is not a perfect square.

In summary, the conversation discusses how to prove that any positive integer with all digits being 1s (except for 1) is not a perfect square. Various methods are suggested, including using a base system and algebraic manipulation. Ultimately, the conversation concludes that the best proof is through contradiction, where assuming a perfect square of this form leads to a logical contradiction.
  • #36
mikeyflex said:
Hi Dick,

The problem stated that p is a prime number that is greater than 3. When you work out p^2(mod 12) you get a remainder of 1 for every prime number square that is greater than 3. This is not the case with 2 or 3. I was wondering if there was some proof behind this.

A direct way to see this is that p(mod 12) can only be 1,5,7 or 11. Otherwise it is divisible by 2 or 3. But 1^2, 5^2, 7^2 and 11^2 are all equal to 1 mod 12. This is what matt was alluding to when he said all primes are equal to +1 or -1 mod 6.
 
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  • #37
matt grime said:
For mikeyflex: every prime not equal to 2 or 3 is of the form 6m+1 or 6m-1 for some integer m. This completely explains your observation.

Thanks Matt,

The problem stated that when p is a prime greater than 3, to evaluate p^2 mod 12.

When you plug in prime values for p^2 mod 12 you get 1, which is an equivalence relation. Is there any significance to this problem as pertaining to a proof of sorts?

Thanks again Matt
 
  • #38
Did you actually read what I wrote and think about it for any length of time. Your observation is completely proven by what I wrote.
 
  • #39
to explain what matt's saying much farther than what needs to be done:
every prime greater than 3 can be written as 6k +1 or 6k -1 (which can be shown easily by a proof by cases).

(6k+1)^2 = 36k^2 + 12k + 1 = 12(3k^2 + k) + 1 = 1mod12
(6k - 1)^2 = 36k^2 -12k +1 = 12(3k^2 - k) + 1 = 1mod12

that's really all the "signifigance" involved in the observation
 
  • #40
matticus said:
to explain what matt's saying much farther than what needs to be done:
every prime greater than 3 can be written as 6k +1 or 6k -1 (which can be shown easily by a proof by cases).

(6k+1)^2 = 36k^2 + 12k + 1 = 12(3k^2 + k) + 1 = 1mod12
(6k - 1)^2 = 36k^2 -12k +1 = 12(3k^2 - k) + 1 = 1mod12

that's really all the "signifigance" involved in the observation


Ahh, now I get it, and that must be the reason why this proof doesn't work for prime numbers 2 and 3. Thank you for the observation. Number theory never seems light the bulb for me.
 

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