46

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 $$(\alpha^2+1)(\beta ^2+1)(\gamma^2+1)(\delta^2+1)=f(i)f(-i)=|f(i)|^2=|-7+3i|^2=49+9=58.$$


27

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 the $w_i$'s are pairwise distinct. Moreover, $$w_1^2 + w_2^2 + \cdots + w_k^2 \equiv \underset{k \: \text{times}}{\underbrace{1 + \cdots + 1}} = k \equiv 0 \pmod{...


26

Lemma 1: $a$ is odd $\Longrightarrow$ $a^2\equiv 1(\operatorname{mod} 4)$. Proof: $a^2-1=(a-1)(a+1)$. Since $a$ is odd, both $a-1$ and $a+1$ are even, so that $a^2-1$ is divisible by $4$. $\blacksquare$ Lemma 2: $a$ is even $\Longrightarrow$ $a^2\equiv 0(\operatorname{mod} 4)$. Proof: Trivial. $\blacksquare$ Now, suppose that $u=a^2+b^2$. (1) If both $a$...


22

I prove it for $n=p$ a prime. After checking the cases $p=2$ and $p=3$ by hand, I suppose that $p>3$. Let $a \in \mathbf F_p$, we want to show that $a$ is the sum of a square and a cube. Without loss of generality we suppose $a \neq 0$. Let $E_a$ be the curve $$y^2 = x^3 +a.$$ over $\mathbf F_p$. Since $a \neq 0$, $E_a$ is an elliptic curve over $\mathbf ...


19

HINT: Using induction: $5^1=2^2+1^2$ Let $5^k=a^2+b^2$ $5^{k+1}=(a^2+b^2)(2^2+1^2)=(2a\pm b)^2+(2b\mp a)^2$ (Brahmagupta–Fibonacci identity) This invites a little generalization: as we which numbers are expressible as the sum of two squares.


18

The original question was to prove that $c\mid a^2+b^2$ implies $c\mid a$ and $c\mid b$, which as many answers show isn't true. But this is true if you take the assumption that there isn't a square root of $-1$ mod $c$. Take mod $c$ of the equation to get that $$a^2 = - b^2 \mod c$$ which can only have a solution $a \neq 0$ or $b \neq 0$ if $-1$ has a ...


18

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 "complete the square" on the left side. Multiply both sides by $8$ and you get: $$16x^2+24x+8=(4x+3)^2-1=48y^2$$ So you need infinitely many solutions to: $$(4x+...


16

There are many counterexamples, of which the smallest is $$5^2 + 5^2 = 7^2 + 1^2.$$ (I earlier stated that $25$ was the smallest counterexample, since $3^2+4^2 = 0^2+5^2$, but in your terminology, it has a potency of 1, not 2, so it is not a counterexample.) Brahmagupta's identity shows that if $x$ and $y$ are each expressible as a sum of two squares, then ...


15

Rabin and Shallit's 1986 paper, "Randomized Algorithms in Number Theory" (Comm. in Pure and Applied Math v.39(supplement):S239-S256), uses "a number of Diophantine problems involving sums of squares" to illustrate efficiencies that are possible with random choices, reducing the expected number of operations while still always producing correct answers (...


15

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 is for $n=2$, so it must be for $n=2+2=4$, etc. Hence by induction, for all even $n \geq 2$ we have that $5^n$ is a sum of two squares. We also know $5^n$ is ...


14

The slickest way is via a little Algebraic Number Theory. If $p=a^2+b^2=c^2+d^2$ then $$p=(a+bi)(a-bi)=(c+di)(c-di)$$ Now ${\bf Z}[i]$ is a unique factorization domain, so these two factorizations of $p$ show that $a+bi$ can't be a prime in ${\bf Z}[i]$. We must have a non-trivial factorization $a+bi=(s+ti)(u+vi)$, whence $a-bi=(s-ti)(u-vi)$, and then $$p=(s^...


14

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 $u < v$, we can write this as $m = (a-c)(a+c)$ where $a = (u+v)/2$ and $c = (v-u)/2$. So any odd number with more than one factorization of this type gives an example. ...


14

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, $$ k(n) = \#\{ 1\le a < \tfrac n2 \colon a^2 + (n-a)^2 = 2a^2 - 2an + n^2 \text{ is prime} \}. $$ Ignoring issues of uniformity in $n$, the Bateman–Horn conjecture predicts that the number of ...


14

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 asymptotically $$ \frac{\sqrt{n}}{\log (n)} \cdot \prod_{p > 2}\left( 1 - \frac{(n / p)}{p - 1}\right)$$ where $(n / p)$ is the Legendre symbol. References: ...


13

$$ 65 = 64 + 1 = 49 + 16 $$ 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.


12

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 proportion of the grid that an $N$-mover can reach is $$f(N)=\left\{\begin{array}{ll} 0&\text{ if } m_j\ \text{ odd for some } j\\ 2^{-k}\prod_jq_j^{-m_j}&\...


11

If: $$x=2ac$$ $$y=2bc$$ $$z=a^2+b^2-c^2$$ $$n=a^2+b^2+c^2$$ Then: $$x^2+y^2+z^2=n^2$$


11

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 this is the case, consider $a= \frac{p+q}{2}, b= \frac{p-q}{2}$. Then, note that $a^2 - b^2 = pq = l$! For example, $183 = 61 \times 3$, so $a=32$ and $b = ...


11

$2n+1=(n+1)^2-n^2+0^2$ and $2n=(n+1)^2-n^2-1^2$ cover the odd and even cases respectively.


11

$$ a^2=b^2+c^2+d^2 \quad\hbox{and}\quad p^2+q^2+r^2=s^2+t^2. $$


11

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 $$3(1^2+2^2+3^2+4^2+5^2)$$ is equal to the area of the rectangle on the left with height $1+2+3+4+5=\frac{6\cdot 5}{2}$ (see the first formula) and base $2\cdot 5+1$. The ...


10

That's not really going to work, the highest degree terms do not cancel here, so all you get is $(2n+1)^2 = (2n+1)^2 + 0^2 + 0^2.$ What does work is due to Gordon Pall; every number is the sum of four squares, so write $$ 2n+1 = a^2 + b^2 + c^2 + d^2.$$ Then you get nontrivial expressions $$ (2n+1)^2 = \left(a^2 + b^2 - c^2 - d^2 \right)^2 + (-2ad+2bc)^2 + (...


10

The problem reduces to, $$k^2+(k+1)^2=(n-1)^2+n^2+(n+1)^2\tag1$$ or simply, $$2k^2+2k-1 = 3n^2\tag2$$ Making a quadratic into a square is a well-studied subject. The complete solution is given by, $$k = \frac{3p^2\pm6pq+4q^2}{3p^2-2q^2}$$ where $3p^2-2q^2=1$. We can transform this condition into a more conventional Pell equation by using the ...


10

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, either $x,y,z$ are all odd, or exactly one of $x,y,z$ is odd. If $x,y,z$ are all odd, then $x^2 + y^2 + z^2 \equiv 3 \pmod 8$, contradiction, since $999 \equiv 7 \...


10

$$ (a-b)^2 + (a+b)^2 = 2(a^2 + b^2) $$ If even $n$ is the sum of two squares, then both are odd or both even; taking $n = c^2 + d^2,$ $$ \frac{n}{2} = \left( \frac{c-d}{2} \right)^2 + \left( \frac{c+d}{2} \right)^2 $$ The rest is induction on the power of 2 dividing you number


10

Using fractional decomposition we get: $$ \frac 1{\left(\frac{k(k+1)}{2}\right)^2} = \frac 4{(k + 1)^2} + \frac 8{(k + 1)} +\frac 4 {k^2} - \frac 8 k $$ So, your sum simplifies to one telescoping sum, and the sum $$\sum_{k=1}^\infty \frac 1 {k^2} =\frac{\pi^2} 6$$ So, all together we get: $$ \frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{6^2}+\frac{1}{10^2}+\dots+\...


10

$$c=\left(\frac{c+1}2\right)^2-\left(\frac{c-1}2\right)^2 =\left(\frac{c+1}2\right)^2+(a_1^2+\cdots+a_n^2)\left(\frac{c-1}2\right)^2.$$


9

No, all numbers can be expressed as the sum of four squares. More information, and a proof using the Hurwitz integers, can be found here.


9

It belongs to a family that starts with the familiar, $$\begin{align} &3^2+4^2 = 5^2\\ &10^2+11^2+12^2=13^2+14^2\\ &21^2+22^2+23^2+24^2 = 25^2+26^2+27^2\\ \vdots\\ &a^2 + (a+1)^2 + \dots + a_n^2 = b^2 + (b+1)^2 + \dots + b_{n-1}^2 \end{align}$$ where $b = a_n+1$ and $a = 2n^2-3n+1 = 0, 3, 10, 21, 36, 55, 78, 105, \color{brown}{136},\dots$ ...


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