ab=cd

Number Theory Level 3

If $ab=cd$

Then $a^2+b^2+c^2+d^2$ is

a,b,c,d belongs to positive integers.

We can't say Always composite Always prime

2 solutions

Max Patrick
Mar 10, 2019

let gcd(a,c) = m so that a a=mn and c=mp. Now let b=pq and d=nq

The sum of squares will now factorise into (m^2 + q^2)(n^2 +p^2) which is always composite.

(In fact "always prime" is very unlikely to be the answer, because no one has ever found a closed formula, permitting an infinite number of input integers, which always outputs a prime number).

What if $m = 1$ and $q = 0$ ?

Jesse Nieminen - 2 years, 3 months ago

not allowed. a,b,c,d are stated to be natural numbers. right at the end there.

Max Patrick - 2 years, 3 months ago

In my experience it is more common that zero is contained in natural numbers.

Jesse Nieminen - 2 years, 3 months ago

Edwin Gray
Mar 10, 2019

we actually show a generalization. Given ab = cd, Let x = g.c.d(a,c) and y = g.c.d(b,d). Then: a = xr, c = xt with (r,t) = 1 and b = yk, d = ym with(k,m) = 1. Substituting, xryk = xtym. Dividing by xy, rk = tm. Since (r,t)= 1, then r|m. Since (k,m) = 1, then m|r. Therefore r = m, and it follows that t = k. a = xr = xm. b = yk = yt. c = xt, and d = ym. Then a^n + b^n + c^n + d^n = (x^n)(m^n) + (y^n)(t^n) + (x^n)(t^n) + (y^n)(m^n) = m^n(x^n + y^n) + t^n(x^n + y^n) = (m^n+ t^n)*(x^n + y^n)

ab=cd

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