Mod Form 493: Fill & Download for Free

[math]\begin {align} (a [/math][math][/math][math]+ b) \[/math][math]mod[/math][math] c &= (a \[/math][math]mod[/math][math] c [/math][math][/math][math]+ b \[/math][math]mod[/math][math] c) \[/math][math]mod[/math][math] c [/math][math][/math][math]\\ (23^{460} [/math][math][/math][math]+ 52^{493}) \[/math][math]mod[/math][math] 11 &= (23^{460} \[/math][math]mod[/math][math] 11 [/math][math][/math][math]\, [/math][math][/math][math]+ \nonumber [/math][math][/math][math]\\ &\qquad [/math][math][/math][math]{} 52^{493} \[/math][math]mod[/math][math] 11) \[/math][math]mod[/math][math] 11 [/math][math][/math][math]\, [/math][math][/math][math]\, \cdots(1) \end{align}[/math]Let us evaluate each term one by one and then substitute back to arrive at the solution.[math]23^{460} \equiv 1^{460} \[/math][math]mod[/math][math] 11 = 1 [/math][math][/math][math]\, [/math][math][/math][math]\, \cdots (2) [/math][math][/math][math][/math][math]52^{493} \equiv 8^{493} \[/math][math]mod[/math][math] 11 [/math][math][/math][math][/math]Now we have to compute [math]8^{493} \[/math][math]mod[/math][math] 11[/math]Using Euler's theorem, we have,[math]8^{10} \[/math][math]mod[/math][math] 11 = 1[/math]Also, [math]8^{490} \[/math][math]mod[/math][math] 11 = 1 [/math][math][/math][math][/math]Therefore, [math]8^{493} \equiv 8^3 \[/math][math]mod[/math][math] 11 [/math][math][/math][math][/math][math]8^{493} \equiv 512 \[/math][math]mod[/math][math] 11[/math][math]512 \[/math][math]mod[/math][math] 11 = 6 [/math][math][/math][math]\, [/math][math][/math][math]\, \cdots (3)[/math]Using [math](2)[/math] and [math](3)[/math] in [math](1)[/math], we get,[math](23^{460} [/math][math][/math][math]+ 52^{493}) \[/math][math]mod[/math][math] 11 = (1 [/math][math][/math][math]+ 6) \[/math][math]mod[/math][math] 11[/math][math](23^{460} [/math][math][/math][math]+ 52^{493}) \[/math][math]mod[/math][math] 11 = 7 \[/math][math]mod[/math][math] 11[/math]Hence the remainder is [math]7[/math].A quick check on Wolfram Alpha verifies my answer.The above screenshot is taken from: Wolfram|Alpha: Making the world’s knowledge computable

Note that every integral primitive Pythagorean triple is generated by the following equations:[math]a=m^2-n^2[/math][math]b=2mn[/math][math]c=m^2+n^2[/math][math]a,b,c[/math] are taken to be the length of the two legs and hypotenuse respectively and [math]m,n [/math][math][/math][math][/math] are arbitrary integers.Given a certain value of [math]c [/math][math][/math][math][/math], we can find [math]m,n [/math][math][/math][math][/math] that satisfy [math]c=m^2+n^2[/math] if and only if the squarefree part of [math]c[/math] is congruent to [math]1[/math] mod [math]4[/math] by Fermat's two square theorem.It is a nontrivial task to actually find [math]c [/math][math][/math][math][/math] as the sum of two squares. A simple way to do this is to take the squarefree part of [math]c[/math] and split it into primes. We can then write these primes as the sum of two squares or look them up in a table. Then we can use an identity of Euler to write the product of the sum of two squares as the sum of two squares:[math](a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2 [/math][math][/math][math][/math]There are probably more efficient ways to find [math]m,n[/math] such that [math]m^2+n^2=c[/math]. At any rate, once [math]m,n[/math] are found, we can plug them into the first equations to find [math]a,b[/math].If that explanation was a bit dense, the following example may help:Consider:[math]c=493[/math][math]493=17 \times 29[/math][math]17=4^2+1^2[/math][math]29=2^2+5^2[/math][math]17\times 29=(8-5)^2+(20+2)^2=3^2+22^2=484+9=493[/math]So we can say that [math]m=22, n=3[/math]. We find that:[math]a=22^2-3^2=475[/math][math]b=132[/math][math](475,132,493)[/math] is therefore a Pythagorean triple.Let's confirm:[math]475^2+132^2=225625+17424=243049=493^2 [/math][math][/math][math][/math]Yup. A Pythagorean triple.

[math]n=11q+9[/math][math]11q+9\equiv 3\,(mod\,5)[/math][math]11q\equiv 4\,(mod\,5)[/math][math]q\equiv 4\,(mod\,5)[/math][math]q=5r+4[/math][math]n=11(5r+4)+9[/math][math]n=55r+53[/math][math]55r+53\equiv 1\,(mod\,3)[/math][math]55r\equiv 2\,(mod\,3)[/math][math]r\equiv 2\,(mod\,3)[/math][math]n=55(2)+53=163[/math][math]n=163+(11)(5)(3)s[/math][math]n=163+165s[/math]For [math]0\leq n \leq 500[/math], there are [math]3[/math] numbers:[math]n=163[/math], [math]n=328[/math], [math]n=493[/math]