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47 votes
Accepted

Value of $(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)$ if $z^4-2z^3+z^2+z-7=0$ for $z=\alpha$, $\beta$, $\gamma$, $\delta$

There is no need to use Vieta's formulas. Let $$f(z)=z^4-2z^3+z^2+z-7=(z-\alpha)(z-\beta)(z-\gamma)(z-\delta).$$ Then, since $(i-a)(-i-a)=-i^2+a^2=1+a^2$, it follows that $$(\alpha^2+1)(\beta ^2+1)(\...
Robert Z's user avatar
  • 146k
39 votes

Algebraically why must a single square root be done on all terms rather than individually?

Note that $\sqrt{x} + \sqrt{y} = \sqrt{x+y} \iff (\sqrt{x} + \sqrt{y})^2 = (\sqrt{x+y})^2 \iff (\sqrt{x} + \sqrt{y})^2 = x + y$. In order to intuitively understand it, let's substitute $$ a = \sqrt{x},...
Arfin's user avatar
  • 1,445
28 votes
Accepted

Can we prove a stronger claim?

Yes. Let $p$ be any prime dividing $k$ (it exists since $k > 1$). Then set $w_1 = p + 1, \: w_2 = 2p + 1, \: \dots, \: w_k = kp + 1$ (in general, $w_i = ip + 1$). Now, clearly $p \nmid w_i$ and ...
cip999's user avatar
  • 1,996
23 votes
Accepted

Is there any variation known to the sum of two squares theorem?

The sum of two squares theorem can be proved by working in $\Bbb Z[i]$. By working in $\Bbb Z[\zeta_3]$ instead, where $\zeta_3$ is a third root of unity, one may prove that an integer greater than ...
Lukas Heger's user avatar
  • 21.7k
22 votes
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Relationship between the squares of first n natural numbers and first n natural odd numbers.

Let $A=\sum_{k=1}^{50} (2k-1)^2$, and $B=\sum_{k=1}^{50} (2k)^2$. Then, $A$ is the answer you want and $A+B=x$. What is $B-A$? Well, We can group that in pairs to get $$B-A=(2^2-1^2)+\cdots (100^2-...
lulu's user avatar
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18 votes
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Find all $x$ so $\frac{1^2+2^2+\cdots+x^2}{x}$ is a perfect square...

The main result is: $$1^2+2^2+3^2+\dots + x^2=\frac{x(x+1)(2x+1)}{6}$$ So you need $\frac{(x+1)(2x+1)}{6}=y^2$ for some $y$. Or: $$2x^2+3x+1 = 6y^2$$ From here, we are essentially going to "...
Thomas Andrews's user avatar
18 votes

Algebraically why must a single square root be done on all terms rather than individually?

An equation is a scale, the kind with two pans, that balances when what's in the one pan weighs the same as what's in the other pan. You can do anything you want to do to one pan, and the scale will ...
Gerry Myerson's user avatar
17 votes

Natural number which can be expressed as sum of two perfect squares in two different ways?

Note that $a^2 + b^2 = c^2 + d^2$ is equivalent to $a^2 - c^2 = d^2 - b^2$, i.e. $(a-c)(a+c) = (d-b)(d+b)$. If we factor any odd number $m$ as $m = uv$, where $u$ and $v$ are both odd and $u < v$, ...
Robert Israel's user avatar
16 votes

Natural number which can be expressed as sum of two perfect squares in two different ways?

$$ 65 = 64 + 1 = 49 + 16 $$ This will work for any number that's the product of two primes each of which is congruent to $1$ mod $4$. For more than two ways multiply more than two such primes.
Ethan Bolker's user avatar
  • 96.4k
16 votes
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Conjecture: All but 21 non-square integers are the sum of a square and a prime

This is Hardy and Littlewood's Conjecture $H$. It says that this sequence $a(n)= 10,34,58,85,\ldots$ is finite and that the number of representations of $n$ as the sum of a prime and a square is ...
Dietrich Burde's user avatar
15 votes
Accepted

How do I prove that powers of 5 are the sum of two squares using mathematical induction

Continuing with your induction, assuming that $5^n=a_n^2+b_n^2$ then multiplying both sides by $5^2$ gives $5^{n+2}=(5a_n)^2+(5b_n)^2$. So $5^{n+2}$ is a sum of two squares if $5^{n}$ is. We know it ...
Ahmed S. Attaalla's user avatar
15 votes

Relationship between the squares of first n natural numbers and first n natural odd numbers.

There is a way to do this without writing any "Sigmas" or calculating the correct result from first principles (or the formula for the sum of squares), but it relies on the question being ...
kaya3's user avatar
  • 1,371
14 votes
Accepted

Can a number be equal to the sum of the squares of its prime divisors?

If $f(n)=n$ then $p_1^{a_1} \cdot ... \cdot p_k^{a_k}=p_1^2+...+p_k^2$. From this it follows that $p_1|p_2^2+...+p_k^2$ and that $p_k|p_1^2+...+p_{k-1}^2$, that is, it is true that $p_2^2+...+p_k^2=...
Grešnik's user avatar
  • 1,802
14 votes
Accepted

Any odd number is of form $a+b$ where $a^2+b^2$ is prime

Here are some heuristics. As Hans Engler defines, let $k(n)$ be the number of pairs $(a,b)$ with $a<b$ for which $a+b=n$ and $a^2+b^2$ is prime. In other words, $$ k(n) = \#\{ 1\le a < \tfrac n2 ...
Greg Martin's user avatar
  • 80.5k
13 votes
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Every number is the sum of three squares with signs

Hang on, it's actually quite simple! So suppose that we have a number $l$. Suppose that $l=pq$, with $p,q$ having the same parity. That is, both $p$ and $q$ are even, or both $p$ and $q$ are odd. If ...
Sarvesh Ravichandran Iyer's user avatar
13 votes
Accepted

Understanding some proofs-without-words for sums of consecutive numbers, consecutive squares, consecutive odd numbers, and consecutive cubes

The second picture gives a visual proof for the formula $$3(1^2+2^2+3^2+\dots +n^2)=\frac{n(n+1)}{2}\cdot (2n+1)$$ for $n=5$. The sum of the areas of the $3\cdot 5$ squares on the right $$3(1^2+2^2+3^...
Robert Z's user avatar
  • 146k
13 votes
Accepted

How to show this inequality $ a^{14}-a^{13}+a^8-a^5+a^2-a+1>0$

Notice that quadratic $x^2-x+1>0$ for all $x$. Now write: $$ \underbrace{(a-1)(a^{13}-1)}_{\geq 0} + a^2(\underbrace{a^6-a^3+1}_{> 0})>0$$
nonuser's user avatar
  • 90.3k
12 votes
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If $n = a^2 + b^2 + c^2$ for positive integers $a$, $b$,$c$, show that there exist positive integers $x$, $y$, $z$ such that $n^2 = x^2 + y^2 + z^2$.

If: $$x=2ac$$ $$y=2bc$$ $$z=a^2+b^2-c^2$$ $$n=a^2+b^2+c^2$$ Then: $$x^2+y^2+z^2=n^2$$
individ's user avatar
  • 4,301
12 votes
Accepted

How much of an infinite board can a N-mover reach?

For a positive integer $N$, write $N=2^k\left(\prod_ip_i^{l_i}\right)\left( \prod_jq_j^{m_j}\right)$, where the $p_i$ and $q_j$ are primes with $p_i\equiv1\pmod{4}$ and $q_j\equiv3\pmod{4}$. The ...
Servaes's user avatar
  • 63.5k
12 votes
Accepted

If $-1$ is a sum of squares, then any element is a sum of squares

$$c=\left(\frac{c+1}2\right)^2-\left(\frac{c-1}2\right)^2 =\left(\frac{c+1}2\right)^2+(a_1^2+\cdots+a_n^2)\left(\frac{c-1}2\right)^2.$$
Angina Seng's user avatar
12 votes

How to show this inequality $ a^{14}-a^{13}+a^8-a^5+a^2-a+1>0$

Notice it a bunch of $+ a^{even}$ and $-a^{odd}$ in descending order of powers so..... Case 1: $a \le 0$ then $a^{even} \ge 0$ and $-a^{odd} \ge 0$ so $a^{14} - a^{13}+a^8-a^5+a^2 -a + 1=|a|^{14}+|a|^...
fleablood's user avatar
  • 125k
11 votes

Prove that there are arbitrarily long sequences of consecutive integers, none of which can be written as the sum of two perfect squares.

A number is a sum of two squares if and only if all primes $3$ mod $4$ appear only to even powers in its prime factorisation. So we want arbitrarily long sequences of consecutive integers such that ...
Patrick Stevens's user avatar
11 votes

Every number is the sum of three squares with signs

$2n+1=(n+1)^2-n^2+0^2$ and $2n=(n+1)^2-n^2-1^2$ cover the odd and even cases respectively.
John Bentin's user avatar
  • 18.8k
11 votes

Does every sum-of-squares equation have a plane geometric interpretation?

$$ a^2=b^2+c^2+d^2 \quad\hbox{and}\quad p^2+q^2+r^2=s^2+t^2. $$
Intelligenti pauca's user avatar
11 votes

Algebraically why must a single square root be done on all terms rather than individually?

A false equation like $\sqrt{9}+\sqrt{16}=\sqrt{9+16}$ is, of course, false, and generally speaking we do not need to go on with elaborate explanations for why evidently false statements are false. ...
Lee Mosher's user avatar
  • 122k
10 votes
Accepted

Sum of Consecutive Perfect Squares

Since $2n + 1$ is odd, if it is a perfect square it is the square of an odd number. So write $2n+1 = (2k+1)^2$. Then we can solve to get $n+1 = \frac{(2k+1)^2 - 1}{2} + 1$, and further simplify this ...
MartianInvader's user avatar
10 votes
Accepted

Let $p$ be a prime so $p\equiv3\pmod4$. If $p|a^2+b^2$, then $p|a,b$

I hope I didn't miss something and I think it is fairly elementary: Using Fermats Little Theorem: $a^p\equiv a\mod{(p)}$ and $b^p\equiv b\mod{(p)}$. Now we get that $a^{p+1}+b^{p+1}\equiv a^2+b^2 \...
CryoDrakon's user avatar
  • 3,422
10 votes

Show that $x^2+y^2+z^2=999$ has no integer solutions

Using congruences . . . Odd squares are always $1 \pmod 8$, hence also $1 \pmod 4$. Even squares are always $0 \pmod 4$, hence either $0 \text{ or } 4 \pmod 8$. Since $x^2 + y^2 + z^2$ is odd, ...
quasi's user avatar
  • 59k
10 votes
Accepted

If N is a sum of two squares, then so is the odd part of N

$$ (a-b)^2 + (a+b)^2 = 2(a^2 + b^2) $$ If even $n$ is the sum of two squares, then both are odd or both even; taking $n = c^2 + d^2,$ $$ \frac{n}{2} = \left( \frac{c-d}{2} \right)^2 + \left( \frac{c+...
Will Jagy's user avatar
  • 140k

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