Testwiki:Reference desk/Archives/Mathematics/2025 February 10

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February 10

Why this equation has no solutions ?

Take $SA$, a semiprime, why is there no solution to this equation

(25² + x×SA)÷(67×y) == (z×SA)²

I’m meaning finding integers x and y and z such as the equation is true and where z≠0 ? 2A01:E0A:401:A7C0:E9E3:F9ED:9B83:6DB8 (talk) 14:36, 10 February 2025 (UTC)

I'm not certain what SA is supposed to be but if it is 1 and z is 1 then it is quite easy to get a solution of x=45 and y=10. NadVolum (talk) 15:24, 10 February 2025 (UTC)

Sorry, edited. $SA$ is semiprime. 2A01:E0A:401:A7C0:C965:8D1D:CEAE:ABA0 (talk) 17:23, 10 February 2025 (UTC)

Multiplying both sides by $67y$ and taking both sides modulo $SA$, we see that $25^2=5^4$ would have to be congruent to $0(\mathrm{mod}SA)$. In other words, $5^4$ would have to be a multiple of a semiprime, which is impossible since it only has one prime factor. GalacticShoe (talk) 17:43, 10 February 2025 (UTC)

and if 25 was replaced by a square which is a multiple of the semiprime, would it works ? In reality 25 can be any perfect square, it doesn’t have to be 25. 2A01:E0A:401:A7C0:C965:8D1D:CEAE:ABA0 (talk) 02:59, 11 February 2025 (UTC)

The equation

$\frac{((m\times \text{SA}{)}^2{)}^2+x\times \text{SA}}{67\times y}=(z\times \text{SA}{)}^2$

is a linear equation in $x.$ It has a unique solution for any choice of non-zero $\text{SA},$ $m,$ non-zero $y,$ and $z.$ If all of these are integers, then so is $x.$  ‑‑Lambiam 06:38, 11 February 2025 (UTC)

With $m$ being a perfect square ? If yes, how to compute it ? 2A01:E0A:401:A7C0:C965:8D1D:CEAE:ABA0 (talk) 13:50, 11 February 2025 (UTC)

You wrote, "if 25 was replaced by a square which is a multiple of the semiprime". Well, $(m\times \text{SA}{)}^2$ is a square that is a multiple of $\text{SA}.$ You don't need to compute $m$; $m$ can be set to any value that you wish, including squares, so you can take $m=100,$ or $m=1,$ or even $m=0.$ Aren't you overdoing the squaring? If $m=a^2$ and you expand the brackets, you end up with ${{a^2}^2}^2{{\text{SA}}^2}^2=a^8{\text{SA}}^4.$  ‑‑Lambiam 21:30, 11 February 2025 (UTC)

Yes, that’s correct, with the ² $m$ is a perfect square by construction. But how to find a value of $m$ such as a solution of $x$ and $y$ and $z$ exists ? 2A01:E0A:401:A7C0:88B1:8A9D:F4E0:D37A (talk) 07:24, 13 February 2025 (UTC)

Take $m=1.$  ‑‑Lambiam 09:27, 13 February 2025 (UTC)