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Prove there do not exist natural numbers $m$ and $n$ such that $7m^2 = n^2$.

Proof: Using the Fundamental Theorem of Arithmetic, we can write $m=(p_1^{r_1 }\ldots p_n^{r_n})$ and $n=(q_1^{s_1 }\ldots q_n^{s_n})$.

Where m and n are natural numbers, $p_i$ and $q_j$ are primes, and $r_i$ and $s_j$ are natural numbers. Then squaring m yields

$m^2=(p_1^{r_1 }\ldots p_n^{r_n })^2$

$m^2=p_1^{2r_1}\ldots p_n^{2r_n}.$

Similarly, by squaring n yields,

$n^2=(q_1^{s_1 }…q_n^{s_n })^2$

$n^2=q_1^{2s_1}…q_n^{2s_n}.$

Now we will have two different cases. Either $p_i = 7$ or $p_i\neq 7$.

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    $\begingroup$ Bookkeeping: what's the number of $7$'s on both sides of the equality. Specifically, is it an even/odd number? $\endgroup$ – Alex R. Oct 22 '14 at 2:31
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The only thing you need to look at in the prime factorizations are the exponents of $7$. Note that $n^2$ will have an even number of sevens in its prime factorization, but $7m^2$ will have an odd number of sevens. This is a contradiction.

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  • $\begingroup$ See also math.stackexchange.com/a/978638/589. $\endgroup$ – lhf Oct 22 '14 at 2:42
  • $\begingroup$ Ok..Could we also say this then?... Case 1: pi ≠ 7 for any i. Then the prime factorization of 7m^2 contains 7^1. Since every exponent in the prime factorization of n2 is a multiple of 7, we see that 7m^2 and n^2 have different prime factorizations. Case 2: pi = 7 for some i. Then the prime factorization of 7m^2 contains 7^(〖2r〗_i+1), and 2^ri+1 is not a multiple of 7. Since every exponent in the prime factorization of n^2 is a multiple of 7, we see that 7m^2 and n^2 have different prime factorizations. Thus in either case, 7m^2 and n^2 have different prime factorizations, so 7m^2≠n^2. $\endgroup$ – Laffabubble Oct 22 '14 at 2:50
  • $\begingroup$ You don't need to break up into cases. You can assume all primes appear in the prime factorization, and allow their exponents to be zero. So case 2 covers both cases. Also, everywhere you said "multiple of 7" you should have said "multiple of 2," but otherwise your reasoning is right. $\endgroup$ – Mike Earnest Oct 22 '14 at 3:05
  • $\begingroup$ Okay great. Thank you so much for your help! $\endgroup$ – Laffabubble Oct 22 '14 at 3:20
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An alternative argument: suppose integers $m$ and $n$ exist such that $7m^2=n^2$. Dividing both sides by common factors of $2$ if necessary, we may assume $m$ and $n$ are both odd, in which case we may write $m=2k+1$ and $n=2l+1$ for integers $k$ and $l$. But then, $$ 1+4(l+l^2)=n^2=7m^2=4(7k^2+7k+1)+3. $$ Taking modulo $4$ of the leftmost and rightmost expression gives the desired contradiction.

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Alternatively, you can prove the square root of 7 is irrational by modifying the famous proof that the square root of 2 is irrational.

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