# Tag Info

143

Note that $x^2\equiv x\pmod 2$ and thus $a^2+b^2=c^2$ implies $$a+b+c\equiv a^2+b^2+c^2\equiv 2c^2\equiv 0\pmod 2$$

41

You cannot have an integer Pythagorean Triple whose angles are $45°, 45°$ and $90°$. Assume on the triangle we have sides $a$. Then by Pythagoras' Theorem, $$a^2+a^2=2a^2=(a\sqrt{2})^2$$ This means the hypotenuse is no longer an integer length, because now it measures $a\sqrt2$. This means no such Pythagoren Triple exists.

40

Hint: If $a,b$ are odd primes, $a^2 + b^2 > 2$ and is even. Hence, the only possibilities are $$2^2 + 2^2 = c^2 \\ 2^2 + b^2 = c^2$$ and they are not possible either because $c - b \ge 2 \implies c^2>2^2 + b^2$.

31

$$\frac{1 + 49}{2} = 25$$ There are infinitely many solutions to $x^2 + y^2 = 2 z^2.$ I am pretty sure that version has been asked on MSE before, and formulas giving all integer answers were given. Give me a few minutes, I will find that or do it over. here is a recent discussion Diophantine Equations : Solving $a^2+ b^2=2c^2$

31

Hint: Suppose you have $A^2+B^2=C^2$ where $A=2a+1, B=2b+1, C=2c$ Substitute, expand and then take out terms with factors of $4$

31

Start from $1105 = 5\cdot 13\cdot 17$ whose prime factors all has the form $4k+1$. Decompose the prime factors in $\mathbb{Z}$ over gaussian integers $\mathbb{Z}[i]$ as $$5 = (2+i)(2-i),\quad 13 = (3+2i)(3-2i)\quad\text{ and }\quad17 = (4+i)(4-i)$$ Recombine the factors of $1105^2$ over $\mathbb{Z}[i]$ in different order and then turn them to sum of two ...

29

Yes, a number can appear as the smallest value in two distinct primitive triples. For example, $(57, 176, 185)$ and $(57, 1624, 1625)$. In fact, choose any two relatively prime positive integers $p$ and $q$ with $q+1 < p < q(1+\sqrt{2})$ and having opposite parity. Then $p^2-q^2 < 2pq$, and $(a,b,c)=(p^2-q^2, 2pq, p^2+q^2)$ is a primitive triple. ...

24

no, but it is pretty elaborate. If there is a solution with positive integers, it is still a solution if we divide out by the greatest common divisor of $a,b.$ We may therefore demand $\gcd(a,b) = 1.$ If $ab$ is a square and they are coprime, both $a,b$ are squares. You are asking for $a = m^2, b = n^2,$ and you want $m^4 + n^4 = c^2.$ This has no nonzero ...

22

Suppose that you are right and there is a possible right triangle such that $a=2k+1, b=2m+1$ and $c=2n$ {legs are odd while hypotenuse is even} and you have $a^2+b^2=c^2$. Now on expanding (substituting for $a, b, c$) you will get that: $$(2k+1)^2+(2m+1)^2=(2n)^2$$ $$4(k^2+m^2+k+m)+2=4n^2$$ On dividing the equation by $2$, you will get $$2(k^2+m^2+k+m)+1=... 20 Also, Pythagorean triples have a well defined structure:$$a=k(m^{2}-n^{2}),\ \,b=k(2mn),\ \,c=k(m^{2}+n^{2})$$and$$a+b+c=2k(mn+m^2)$$19 Let's suppose that a^{-n}+b^{-n} = c^{-n} for some positive integers a,b,c,n. Then, multiply both sides by a^nb^nc^n to get b^nc^n + a^nc^n = a^nb^n, i.e. (bc)^n+(ac)^n=(ab)^n. If n \ge 3, then this contradicts Fermat's Last Theorem. Hence, there are no solutions for n \ge 3. For n = 1, we have several solutions, one of which is 3^{-1}+... 19 Hint Write a+b+c=k, so$$a^2+b^2=(a+b)^2-2ab= (k-c)^2-2ab=c^2 → k^2-2(kc+ab)=0→k^2=2(kc+ab)$$18 Notice that (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)=2c^2+2(ab+bc+ac), so the square of a+b+c is even and thus a+b+c is also even. 17 There is an infinite number of such pythagorean triples. Any primitive pythagorean triple (a,b,c) with a^2+b^2=c^2 (are we are looking for primitive triples since we want a and b consecutive) is of the form:$$ a=p^2-q^2,\qquad b=2pq,\qquad c=p^2+q^2 $$with p and q coprime and not both odd. So we are looking for integer solutions of:$$ p^2-2pq-...

17

We have $a^2 + p^2 = b^2$ so $p^2 = b^2-a^2 = (b+a)(b-a)$. Therefore $b-a=1$ and $p^2 = b+a=2a+1$ Therefore $2(a+p+1) = p^2 + 2p + 1 = (p+1)^2$

16

From $a^2+b^2=c^2$ we get $a^2=c^2-b^2=(c+b)(c-b)$, i.e. a factorization of $a^2$ into two distinct factors $c+b>c-b$. The only such factorizations for the square of a prime is $a^2\cdot 1$, i.e. we conclude $c-b=1$, hence $b=2$, $c=3$. But then $a^2=5$, qea.

16

Parametric solution: $$a = |x^2 - 2 x y - y^2|, \ b = x^2 + y^2,\ c = x^2 + 2 x y - y^2$$

16

The "smallest" counterexample is \begin{gather*} 5^2 + 12^2 = 13^2 \\ 9^2 + 12^2 = 15^2 \\ 12^2 + 16^2 = 20^2 \end{gather*} ($12$ also satisfies $12^2 + 35^2 = 37^2$.) Edit: now that the OP has stipulated that the elements be pairwise coprime, the smallest counterexample (in the sense that the repeated number is minimal) is \begin{gather*} 11^2 + 60^2 = 61^...

15

As noted in the comments and the accepted answer, this comes down to the fact that if a prime $p$ can be written as a sum of two squares, then the representation is unique up to switching and or negating the factors. A fancier explanation for this is the fact that $\mathbb Z[i]$ is a principal ideal domain and its unit group is $\{\pm1,\pm i\}$. (Of course, ...

15

The example $(3,4,5)$ and $(5,12,13)$ shows that some positive integers can appear in more than one primitive triple.

15

The solution $(x,y,z)=(0,1,1)$ works for all $n$. If you don't want to allow $0$, then let $x,y\in\Bbb{N}$ be such that $$x+yi=(1+2i)^n.$$ Then $$5^n=((1+2i)(1-2i))^n=(1+2i)^n(1-2i)^n=(x+yi)(x-yi)=x^2+y^2.$$ Alternatively, if $n$ is odd let $m:=\frac{n-1}{2}$ so that $$(2^m)^2+(2^m)^2=2^n.$$ Finally, less constructively, a theorem of Gauss tells us that ...

15

No, since if the perpendicular sides are $a$ in length, the hypotenuse would be $a\sqrt2$. But $\sqrt2$ is irrational, so $a\sqrt2$ is not an integer.

14

The hypotenuse of a primitive triple is never divisible by $3$. For let $(x,y,z)$ be a primitive triple, and suppose $3$ divides $z$. Then $3$ cannot divide $x$ or $y$, else the triple would not be primitive. It follows that $x^2$ and $y^2$ have remainder $1$ on division by $3$, which means $x^2+y^2$ has remainder $2$ on division by $3$. Remark: A ...

14

If $n$ is odd, we have $$n^2 + \left(\dfrac{n^2-1}{2}\right)^2 = \left(\dfrac{n^2+1}{2}\right)^2$$ If $n$ is even we have $$n^2 + \left( \dfrac{n^2}{4}-1 \right)^2 = \left( \dfrac{n^2}{4} + 1 \right)^2$$

14

Let the legs measure $a$ and $ka$ with $k>1$. Then the approximation given is: $\frac{7ka}{8}+\frac{4a}{8}=\frac{(7k+4)a}{8}$. The real measure is $\sqrt{k^2a^2+a^2}=\sqrt{k^2+1}a$. So how does $\frac{7k+4}{8\sqrt{k^2+1}}$ behave? Not that bad. Here is a graph: So as the graph shows, in the interval $(1,\infty)$ the best approximation occurs at $k=1$ ...

14

This goes back to Euler, who showed that if there are two ways of writing an odd integer $N$ as the sum of two squares, then $N$ is composite. There is a 2009 article on this by Brillhart. Let me try to find a link. http://www.maa.org/press/periodicals/american-mathematical-monthly/american-mathematical-monthly-december-2009 And if one note that in a ...

14

Yes, it does revolve around the fact that $\sqrt{2}$ is irrational. For if there were integers $a$ and $c$ such that $a^2 + a^2 = c^2$, then $2a^2 = c^2$, or $2 = \left(\frac{c}{a}\right)^2$. Therefore $\sqrt{2}$ is rational $\Rightarrow\Leftarrow$.

14

In your context you might be interested in isosceles triangles that are almost right. As others have said, a right isosceles triangle has sides that are $a,a,a\sqrt 2$ and as $\sqrt 2$ is not rational we cannot have an integer sided one. However, if we find a rational number that is close to $\sqrt 2$ we can find isosceles triangles that are close to right....

13

Divide by $c^x$, and get $$\left(\frac{a}{c}\right)^x+\left(\frac{b}{c}\right)^x=1$$ The left-hand side is a decreasing function of $x$. It equals 2 when $x=0$ and approaches 0 for large $x$, so there will be only one solution. It is greater than $2$ when $x<0$ because $a$ and $b$ are less than $c$.

13

Pythagorean triplets are characterized by being of the form $$a= m^2-n^2 \\ b= 2mn \\ c=m^2+n^2$$ So you could look for two integers $m>n$ such that $$\frac{m^2-n^2}{2mn} \approx \frac{5}{7}$$ or, equivalently, $$\frac{m}{n}- \frac{n}{m} \approx 2 \cdot \frac{5}{7}$$ Now, let $x$ be unique positive solution of $$x-x^{-1} = 2 \cdot \frac{5}{7}$$ one can ...

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