ch2.0.5-impossibility-quadratic-squares

Ch2.0.5 — Of the Cases in which the Formula $a+bx+c{x}^2$ can never become a Square

Summary: Uses residue-class analysis modulo 3, 4, 5, and 7 to identify infinite families of two-term formulas $a{t}^2+b{u}^2$ that can never be perfect squares, providing a practical impossibility test before attempting Diophantine solutions.

Sources: chapter-2.0.5

Last updated: 2026-05-08

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Reduction to Two Terms (§63)

Any three-term formula $a+bx+c{x}^2$ can be stripped of its middle term by the substitution $x=(y-b)/(2c)$, which transforms it into

$\frac{4ac-b^2+y^2}{4c}.$

Setting this equal to $z^2/4$ gives $y^2=c{z}^2+b^2-4ac$. Writing $t=b^2-4ac$, the original formula is a square if and only if $c{z}^2+t$ is a square. Because this two-term form is easier to analyse, all impossibility arguments reduce to it.

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Mod-3 Analysis (§65–68)

Every integer belongs to one of three residue classes $3n$, $3n+1$, $3n+2$. Computing squares:

Class	Square	Mod-3 remainder
$3n$	$9n^2$	$0$
$3n+1$	$9n^2+6n+1$	$1$
$3n+2$	$9n^2+12n+4$	$1$

Conclusion: squares mod 3 are only $0$ or $1$; residue $2$ is impossible for a square.

Application (§66): $3x^2+2$ can never be a square, for any integer or fraction $x$.

• If $x$ is an integer: dividing $3x^2+2$ by 3 leaves remainder 2.
• If $x=t/u$ in lowest terms: clearing denominators gives $3t^2+2u^2$. If $u$ is divisible by 3 then $t$ is not, and $3t^2+2u^2$ is divisible by 3 but not 9. If $u$ is not divisible by 3 then $u^2\equiv 1(\mathrm{mod}3)$, so $2u^2\equiv 2(\mathrm{mod}3)$ and the whole expression has remainder 2.

Generalization (§67–68): The argument extends to $3t^2+(3n+2)u^2$ for all integers $n$ (including negative), proving that formulas whose second coefficient is $\equiv 2(\mathrm{mod}3)$ are never squares.

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Mod-4/8 Analysis (§69–71)

Every number belongs to one of $4n$, $4n+1$, $4n+2$, $4n+3$. Squares:

Class	Square	Remainder
$4n$	$16n^2$	$0(\mathrm{mod}16)$
$4n+1$	$16n^2+8n+1$	$1(\mathrm{mod}8)$
$4n+2$	$16n^2+16n+4$	$4(\mathrm{mod}16)$
$4n+3$	$16n^2+24n+9$	$1(\mathrm{mod}8)$

Conclusions (§70):

• Even squares are of the form $16n$ or $16n+4$; forms $16n+2$, $16n+6$, $16n+8$, $16n+10$, $16n+12$, $16n+14$ are never squares.
• Odd squares are all $\equiv 1(\mathrm{mod}8)$; forms $8n+3$, $8n+5$, $8n+7$ are never squares.

Second impossibility proof for $3t^2+2u^2$ (§71): $t$ and $u$ cannot both be even (they share a divisor 2, contradicting the lowest-terms assumption). If both odd: $3t^2\equiv 3(\mathrm{mod}8)$ and $2u^2\equiv 2(\mathrm{mod}8)$, giving remainder $5$ — not a square. If $t$ even and $u$ odd: $3t^2\equiv 0(\mathrm{mod}4)$ and $2u^2\equiv 2(\mathrm{mod}4)$, remainder $2$ — not a square. Similarly for $t$ odd and $u$ even. Hence $3t^2+2u^2$ is never a square.

The same argument proves that $(8m+3)t^2+(8n+2)u^2$ can never be a square for any integers $m$, $n$.

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Mod-5 Analysis (§72–73)

Numbers fall into five classes mod 5. Squares of non-multiples of 5:

Class	Square mod 5
$5n+1$	$1$
$5n+2$	$4$
$5n+3$	$4$
$5n+4$	$1$

Non-zero squares mod 5 are only $1$ or $4$; residues $2$ and $3$ are impossible.

Application (§73): Neither $5t^2+2u^2$ nor $5t^2+3u^2$ can ever be a square. If $u$ is divisible by 5, these are divisible by 5 but not 25, so not squares. If $u^2\equiv 1(\mathrm{mod}5)$: $5t^2+2u^2$ leaves $2(\mathrm{mod}5)$ and $5t^2+3u^2$ leaves $3(\mathrm{mod}5)$. If $u^2\equiv 4(\mathrm{mod}5)$: the remainders swap (3 and 2 respectively). In all cases, neither expression is a square.

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General Divisor $d$ (§75–77)

For any divisor $d$, the residue classes of squares satisfy the symmetry: $dn+a$ and $dn-a$ both give remainder $a^2\mathrm{mod}d$ when squared. Thus the distinct square residues mod $d$ correspond to $0,1^2,2^2,\dots$ taken mod $d$.

Mod-7 table (§75–76):

Class	Square	Remainder mod 7
$7n$	—	$0$
$7n\pm 1$	—	$1$
$7n\pm 2$	—	$4$
$7n\pm 3$	—	$2$

Non-zero squares mod 7 are only $1$, $2$, $4$; residues $3$, $5$, $6$ are impossible.

Application (§78): $7t^2+3u^2$, $7t^2+5u^2$, and $7t^2+6u^2$ can never be squares; dividing $u^2$ by 7 gives only remainders $1$, $2$, $4$, but multiplying by $3$, $5$, or $6$ always leaves residues outside $\{0,1,2,4\}$.

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Combined Criterion (§74)

Since all odd squares are $\equiv 1(\mathrm{mod}4)$, the general formula

$(4m+3)t^2+(4n+3)u^2$

can never be a square: if both $t$ and $u$ are odd, the remainder is $2(\mathrm{mod}4)$; if one is even, the remainder is $3(\mathrm{mod}4)$. Neither 2 nor 3 is a square residue mod 4.

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Practical Summary

Before attempting to find integer solutions to $a+bx+c{x}^2=\square$:

1. Reduce to the two-term form $c{t}^2+s{u}^2$ by the substitution of §63.
2. Check whether this form can leave only non-square residues modulo some divisor $d$.
3. If so, the formula is permanently impossible — no rational or integer solution exists.
4. If not, a single solution (if found) generates infinitely many via the method of ch2.0.6-integer-solutions-quadratic-squares.

The quadratic-residues concept page collects the full residue tables used here.

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Related pages

• ch2.0.4-surd-rationalization
• ch2.0.6-integer-solutions-quadratic-squares
• quadratic-residues
• indeterminate-analysis
• rationalization
• discriminant