# Thread: X, Y, Z are coprime integers. 1/x + 1/y = 1/z. Prove that (X+Y) is a square number.

1. ## X, Y, Z are coprime integers. 1/x + 1/y = 1/z. Prove that (X+Y) is a square number.

Okay, so I'm stuck on this for an hour now, so I'd really appreciate some help, please:

X, Y, Z are coprime integers. We also know that 1/x + 1/y = 1/z. Prove that (X+Y) is a square number.

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This should be entry level university math, but I'm stuck with it. I hope I posted to the right topic. Thanks for the help.

2. ## Re: X, Y, Z are coprime integers. 1/x + 1/y = 1/z. Prove that (X+Y) is a square numbe

Are x, y, z pairwise coprime or coprime together?

In mathematics, the case of variable names matters. For example, X and x are different variables.

3. ## Re: X, Y, Z are coprime integers. 1/x + 1/y = 1/z. Prove that (X+Y) is a square numbe

I apologize. Please consider x=X and so on. My mistake.

They are coprime together: The problem says that their greatest common divisor is 1.

Thank you.

4. ## Re: X, Y, Z are coprime integers. 1/x + 1/y = 1/z. Prove that (X+Y) is a square numbe

OK, here is a sketch. Suppose that p, q, r are prime, i, j₁, j₂ and k are positive integers, and $x=p^iq^{j_1}$, $y=q^{j_2}r^k$. In the general case, the variables here should be vectors. Without loss of generality, let $j_1\le j_2$. Then

$z=\frac{xy}{x+y} =\frac{p^iq^{j_1+j_2}r^k}{q^{j_1}(p^i+q^{j_2-j_1}r^k)} =\frac{p^iq^{j_2}r^k}{p^i+q^{j_2-j_1}r^k}$.

Let $s=p^i+q^{j_2-j_1}r^k}$. Then the only primes that could divide s are p, q and r. If p | s, then $p\mid q^{j_2-j_1}r^k$, a contradiction. Similarly, r | s is false. Thus $s=q^j$ for some j. Moreover, $j=j_2$; otherwise q ∈ GCD(x, y, z). Finally, if $j_2>j_1$, then q | s implies $q\mid p^i$, a contradiction. Therefore, $j_1=j_2=j$, $s=q^j$ and $x+y=q^{2j}$.

See if you can shorten this and if it can be extended to the general case.

5. ## Re: X, Y, Z are coprime integers. 1/x + 1/y = 1/z. Prove that (X+Y) is a square numbe

Originally Posted by emakarov
OK, here is a sketch. Suppose that p, q, r are prime, i, j₁, j₂ and k are positive integers, and $x=p^iq^{j_1}$, $y=q^{j_2}r^k$. In the general case, the variables here should be vectors. Without loss of generality, let $j_1\le j_2$. Then

$z=\frac{xy}{x+y} =\frac{p^iq^{j_1+j_2}r^k}{q^{j_1}(p^i+q^{j_2-j_1}r^k)} =\frac{p^iq^{j_2}r^k}{p^i+q^{j_2-j_1}r^k}$.

Let $s=p^i+q^{j_2-j_1}r^k}$. Then the only primes that could divide s are p, q and r. If p | s, then $p\mid q^{j_2-j_1}r^k$, a contradiction. Similarly, r | s is false. Thus $s=q^j$ for some j. Moreover, $j=j_2$; otherwise q ∈ GCD(x, y, z). Finally, if $j_2>j_1$, then q | s implies $q\mid p^i$, a contradiction. Therefore, $j_1=j_2=j$, $s=q^j$ and $x+y=q^{2j}$.

See if you can shorten this and if it can be extended to the general case.
To extend your ideas to the general case, may I suggest that we write x = da and y = db where d = gcd(x,y).
From this we get dab=(a+b)z and using your method we show that d = a + b.

6. ## Re: X, Y, Z are coprime integers. 1/x + 1/y = 1/z. Prove that (X+Y) is a square numbe

Generally this equation quotient more general equation.

equation: $XY+XZ+YZ=N$

Solutions in integers can be written by expanding the number of factorization: $N=ab$

And vospolzovavschis solutions of Pell's equation: $p^2-(4k^2+1)s^2=1$

$k$ - what some integer number given by us.

Solutions can be written:

$X=ap^2+2(ak+b+a)ps+(2(a-2b)k+2b+a)s^2$

$Y=2(ak-b)ps+2(2ak^2+(a+2b)k+b)s^2$

$Z=bp^2-2(2b+a)kps+(4bk^2-2ak-b)s^2$

And more:

$X=-2bp^2+2(k(4b+a)+b)ps-2((4b+2a)k^2+(2b-a)k)s^2$

$Y=-(2b+a)p^2+2(k(4b+a)-b-a)ps-(8bk^2-(4b+2a)k+a)s^2$

$Z=bp^2-2(2b+a)kps+(4bk^2-2ak-b)s^2$

Perhaps these formulas for someone too complicated. Then equation: $XY+XZ+YZ=N$

If we ask what ever number: $p$ That the following sum can always be factored: $p^2+N=ks$

Solutions can be written.

$X=p$

$Y=s-p$

$Z=k-p$