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Does there exist $a \in \mathbb Z$ such that $a^2 = 5,158, 232,468,953,153$? Use the work from part a to justify the answer.

So in part a, I had to prove:

For each $[a] \in \mathbb Z_5 ,$ if $[a] \neq [0] $, then $[a]^2 = [1]$ or $[a]^2=[4]$.

I was able to this, and determined that this was true for all of the cases.

Using the information I found from that part of the question though, I have to prove if there exists $a \in \mathbb Z$ such that $a^2 = 5,158, 232,468,953,153$.

All that I have been able to write down for my answer so far is: For each integer $a$, if $\ a \ncong 0 (\mod5)$, then $a^2 \cong 1 (\mod5)$ or $ a^2 \cong 4 (\mod5)$ and integer $n$ is a perfect square if $\exists k \in \mathbb Z : n=k^2.$

I'm not sure where to go from here, any help would be appreciated!

Robert Howard
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Claire
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2 Answers2

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So start by writing down the squares of the integers from $1,\ldots,10$. Note that they end in only one of $\{1,4,6,9\}$. It is also easy to see that the last digit of a perfect square depends only on the last digit of its square root (write the square root as $10k+a$). You'll then see that any perfect square ends in one of the above digits, and 3 is not one of them. It also follows from the fact that you already found out that a perfect square is congruent module $1$ or $4$ to 5, which is essentially the same statement. Clearly, any integer which has $3$ as its last digit is congruent modulo $3$, which is a contradiction. PS: integers which are multiples of $10$ end with a $0$ (in fact, two of them), but I assume you aren't considering squares of $0$ when you look at $\mathbb Z_5$.

Boshu
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You yourself wrote down

For each integer a,if a≆0 (mod 5),then a^2≅1(mod 5) or a^2≅4(mod 5)

Does $a^2 = 5,158,232,468,953,153$ satisfy those conditions?

However you have to consider that maybe $a \cong 0 \pmod 5$. In which case $a^2 \cong 0 \pmod 5$. Do $a^2$ satisfy that condition either?

fleablood
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  • No, because either way, the number doesn't end in a 0 or a 5, so it's not divisible by 5. The 0 condition isn't satisified either because perfect squares only end in 1, 4, 6, or 9 – Claire Nov 30 '18 at 02:59
  • Is this the reason why it's not true then? @fleablood – Claire Nov 30 '18 at 03:18
  • Well, you DID say $a^2$ must be equiv $1$ or $4$ mod $5$. Or $a$ equiv $0$. And $5,1....,.53$ is not that. So you tell me. Is it possible for something to happen that you yourself proved could not happen? – fleablood Nov 30 '18 at 04:27
  • But perfect squares like 100, 400, 900, 1600, etc. can and with $0$. – fleablood Nov 30 '18 at 04:28